Again Applying Fubinis Theorem but This Time Integrating With Respect to X First We Have

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Consider a surface \(f(x,y)\text{.}\) In this department, we are interested in calculating either the book under \(f\) or the average function value of \(f\) over a certain area in the \(ten\)-\(y\)-aeroplane. Yous might temporarily think of this surface every bit representing physical topography—a hilly landscape, perhaps. What is the average peak of the surface (or average altitude of the mural) over some region?

Subsection 4.2.ane Book and Average Value over a Rectangular Region

As with most such issues, we beginning by thinking about how we might approximate the answer. Suppose the region is a rectangle, \([a,b]\times[c,d]\text{.}\) Nosotros can divide the rectangle into a grid, \(m\) subdivisions in ane direction and \(n\) in the other, as indicated in the 2D graph below, where \(m=6\) and \(n=4\text{.}\) We pick \(x\)-values \(x_0\text{,}\) \(x_1\text{,}\)…, \(x_{m-1}\) in each subdivision in the \(10\)-management, and similarly in the \(y\)-direction. At each of the points \((x_i,y_j)\) in ane of the smaller rectangles in the grid, nosotros compute the height of the surface: \(f(x_i,y_j)\text{.}\) Now the average of these heights should be (depending on the fineness of the grid) close to the average height of the surface:

\begin{equation*} \ds\frac{f(x_0,y_0)+f(x_1,y_0)+\cdots+f(x_0,y_1)+f(x_1,y_1)+\cdots+ f(x_{m-one},y_{n-1})}{mn} \finish{equation*}

As both \(m\) and \(n\) go to infinity, we expect this approximation to converge to a fixed value, the actual boilerplate height of the surface. For reasonably nice functions this does indeed happen.

Using sigma note, we can rewrite the approximation:

\begin{align*} \frac{1}{mn}\sum_{i=0}^{due north-ane}\sum_{j=0}^{m-1}f(x_j,y_i) \amp =\frac{one}{(b-a)(d-c)}\sum_{i=0}^{n-one}\sum_{j=0}^{thou-ane}f(x_j,y_i)\frac{b-a}{thou}\frac{d-c}{northward}\\ \amp =\frac{ane}{(b-a)(d-c)}\sum_{i=0}^{due north-1}\sum_{j=0}^{g-1}f(x_j,y_i)\Delta ten\Delta y\text{.} \cease{align*}

The two parts of this product have useful significant: \((b-a)(d-c)\) is of course the area of the rectangle, and the double sum adds upwardly \(mn\) terms of the class \(f(x_j,y_i)\Delta x\Delta y\text{,}\) which is the meridian of the surface at a indicate multiplied by the area of i of the small rectangles into which we take divided the large rectangle. In short, each term \(f(x_j,y_i)\Delta ten\Delta y\) is the volume of a alpine, sparse, rectangular box, and is approximately the volume under the surface and above one of the small rectangles; come across Figure 4.1. When we add all of these up, we become an approximation to the book under the surface and above the rectangle \(R=[a,b]\times[c,d]\text{.}\) When nosotros have the limit every bit \(m\) and \(n\) go to infinity, the double sum becomes the actual volume under the surface, which nosotros divide past \((b-a)(d-c)\) to go the average height.

(a) The surface \(z\) on \([0.5,3.5]\times[0.v,2.5]\text{.}\)

(b) Approximating the book under \(z\text{.}\)

Figure 4.i. The surface \(z=\sin(xy)+\frac{six}{5}\text{.}\)

Double sums similar this come up in many applications, then in a way it is the most of import part of this example; dividing past \((b-a)(d-c)\) is a simple extra pace that allows the computation of an average. This computation is independent of \(f\) being positive, and so every bit we did in the single variable instance, nosotros innovate a special notation for the limit of such a double sum, which is referred to as the double integral of \(f\) over the region \(R\text{.}\)

Definition 4.11. Double Integral over a Rectangular Region.

Permit \(f(10,y)\) be a continuous function defined over the rectangle \(R=[a,b]\times[c,d]\text{,}\) then the double integral of \(f\) over \(R\) is denoted past

\brainstorm{equation*} \iint_R f(x,y)\,dA=\iint_R f(10,y)\,dx\,dy=\lim_{thou\to\infty}\lim_{n\to\infty} \sum_{i=0}^{due north-ane}\sum_{j=0}^{thousand-1}f(x_j,y_i)\Delta x\Delta y\text{,} \stop{equation*}

provided the limit exists.

Annotation: The annotation \(dA\) indicates a small element of area, without specifying whatever particular order for the variables \(x\) and \(y\text{;}\) it is shorter and more than generic than writing \(dx\,dy\text{.}\)

We now capture our results from the before calculations using the notation of the double integral.

Theorem 4.12. Boilerplate Value of a Two-variable Office.

Let \(f(x,y)\) be a continuous function defined over the rectangle \(R=[a,b]\times[c,d]\text{,}\) then the \thmfont{average value \(f_{avg}\)} of \(f\) over \(R\) is

\begin{equation*} f_{avg} = \frac{1}{(b-a)(d-c)} \iint_R f(10,y)\,dA\text{,} \end{equation*}

provided the double integral exists.

Theorem iv.13. Volume Beneath a Surface.

Let \(f(x,y)\) be a continuous office defined over the rectangle \(R=[a,b]\times[c,d]\text{,}\) with \(f(x,y) \geq 0\text{,}\) and so the \thmfont{volume \(V\) below the surface} \(f\) is

\begin{equation*} 5 = \iint_R f(x,y)\,dA\text{,} \end{equation*}

provided the double integral exists.

Note: Only as with unmarried-variable integration, this last theorem can be stated for more general functions that tin can take on negative values, as long as we understand that \(V\) represents a signed volume.

Subsection 4.2.2 Computing Double Integrals over Rectangular Regions

The side by side question, of class, is: How do we compute these double integrals? Yous might think that we will need some two-dimensional version of the Cardinal Theorem of Calculus, only equally it turns out we can go away with only the unmarried variable version, applied twice.

Going back to the double sum, we can rewrite information technology to emphasize a particular social club in which we desire to add the terms:

\brainstorm{equation*} \sum_{i=0}^{n-one}\left(\sum_{j=0}^{m-1}f(x_j,y_i)\Delta ten\right)\Delta y\text{.} \cease{equation*}

In the sum in parentheses, but the value of \(x_j\) is changing; \(y_i\) is temporarily constant. As \(m\) goes to infinity, this sum has the right form to plough into an integral:

\begin{equation*} \lim_{m\to\infty}\sum_{j=0}^{m-ane}f(x_j,y_i)\Delta ten = \int_a^b f(x,y_i)\,dx\text{.} \end{equation*}

Then later we accept the limit as \(m\) goes to infinity, the sum is

\begin{equation*} \sum_{i=0}^{n-1}\left(\int_a^b f(x,y_i)\,dx\right)\Delta y\text{.} \cease{equation*}

Of course, for dissimilar values of \(y_i\) this integral has different values; in other words, it is really a function applied to \(y_i\text{:}\)

\brainstorm{equation*} Chiliad(y)=\int_a^b f(10,y)\,dx\text{.} \end{equation*}

If we substitute back into the sum we become

\begin{equation*} \sum_{i=0}^{due north-1} M(y_i)\Delta y\text{.} \end{equation*}

This sum has a nice estimation. The value \(Chiliad(y_i)\) is the area of a cross section of the region nether the surface \(f(x,y)\text{,}\) namely, when \(y=y_i\text{.}\) The quantity \(G(y_i)\Delta y\) can be interpreted as the volume of a solid with face area \(G(y_i)\) and thickness \(\Delta y\text{.}\) Think of the surface \(f(ten,y)\) as the meridian of a loaf of sliced bread. Each piece has a cantankerous-sectional surface area and a thickness; \(Yard(y_i)\Delta y\) corresponds to the volume of a single slice of bread. Adding these upward approximates the total book of the loaf. (This is very similar to the technique we used to compute volumes in Section 3.3, except that there we need the cross-sections to be in some way "the aforementioned" .) Figure 4.2 shows this "sliced loaf" approximation using the aforementioned surface as shown in Figure 4.ane. Nicely enough, this sum looks just like the sort of sum that turns into an integral, namely,

\begin{marshal*} \lim_{n\to\infty}\sum_{i=0}^{n-ane} G(y_i)\Delta y\amp =\int_c^d G(y)\,dy\\ \amp =\int_c^d \int_a^b f(ten,y)\,dx\,dy\text{.} \end{align*}

Permit's be clear near what this means: we outset will compute the inner integral, temporarily treating \(y\) as a constant. We will do this by finding an antiderivative with respect to \(10\text{,}\) then substituting \(10=a\) and \(10=b\) and subtracting, every bit usual. The result volition be an expression with no \(10\) variable but some occurrences of \(y\text{.}\) Then the outer integral will be an ordinary i-variable trouble, with \(y\) equally the variable.

Figure 4.2. Approximating the volume under a surface with slices.

Example 4.fourteen. Boilerplate Acme.

Figure four.1 shows the function \(\sin(xy)+6/5\) on \([0.5,3.five]\times[0.5,2.5]\text{.}\) Detect the average superlative of this surface.

Solution

The volume under this surface is

\begin{equation*} \int_{0.5}^{2.five}\int_{0.5}^{3.5} \sin(xy)+{6\over5}\,dx\,dy\text{.} \end{equation*}

The inner integral is

\brainstorm{equation*} \begin{carve up} \int_{0.five}^{3.5} \sin(xy)+{6\over5}\,dx\amp= \left.{-\cos(xy)\over y}+{6x\over5}\right|_{0.5}^{3.five}\\ \amp= {-\cos(iii.5y)\over y}+{\cos(0.5y)\over y}+{xviii\over5}. \stop{carve up} \end{equation*}

Unfortunately, this gives a function for which we tin't find a simple antiderivative. To complete the problem we could use Sage or similar software to approximate the integral. Doing this gives a volume of approximately \(8.84\text{,}\) so the average height is approximately \(8.84/6\approx one.47\text{.}\)

Because addition and multiplication are commutative and associative, we can rewrite the original double sum:

\begin{equation*} \sum_{i=0}^{n-1}\sum_{j=0}^{grand-1}f(x_j,y_i)\Delta 10\Delta y=\sum_{j=0}^{m-1}\sum_{i=0}^{n-ane}f(x_j,y_i)\Delta y\Delta x\text{.} \cease{equation*}

At present if we echo the development to a higher place, the inner sum turns into an integral:

\begin{equation*} \lim_{n\to\infty}\sum_{i=0}^{n-1}f(x_j,y_i)\Delta y = \int_c^d f(x_j,y)\,dy\text{,} \finish{equation*}

and then the outer sum turns into an integral:

\begin{equation*} \lim_{m\to\infty}\sum_{j=0}^{m-ane}\left(\int_c^d f(x_j,y)\,dy \right)\Delta x = \int_a^b\int_c^d f(x,y)\,dy\,dx\text{.} \finish{equation*}

In other words, nosotros can compute the integrals in either lodge, first with respect to \(ten\) then \(y\text{,}\) or vice versa. Thinking of the loaf of bread, this corresponds to slicing the loaf in a management perpendicular to the first.

We summarize our findings past providing a general guideline for how the double integral over a rectangle, such equally the one shown below, is computed.

Guideline for Computing a Double Integral over a Rectangle.

Let \(f(10,y)\) be a continuous office defined over the rectangle \(R=[a,b]\times[c,d]\text{.}\) Follow these steps to evaluate the double integral

\begin{equation*} \int_c^d \int_a^b f(10,y)\,dx\,dy = \int_a^b \int_c^d f(x,y)\,dy\,dx\text{.} \end{equation*}

1. Choose the double integral, \(\ds{\int_c^d \int_a^b f(x,y)\,dx\,dy}\) or \(\ds{\int_a^b \int_c^d f(x,y)\,dy\,dx}\text{.}\)

2. Compute the inner integral:

   1. If \(\ds \int_a^b f(x,y)\,dx\text{,}\) then temporarily treat \(y\) as a constant. Detect an antiderivative with respect to \(x\text{,}\) then evaluate over the integration premises \(x = a\) and \(x = b\text{.}\)

   2. If \(\ds \int_c^d f(x,y)\,dy\text{,}\) then temporarily care for \(x\) equally a constant. Find an antiderivative with respect to \(y\text{,}\) then evaluate over the integration bounds \(y=c\) and \(y=d\text{.}\)

3. Compute the outer integral, which will exist an ordinary one-variable integral of the form

   \begin{equation*} \int_c^d Thou(y)\, dy \text{ or } \int_a^b 1000(10)\,dx\text{.} \end{equation*}

We haven't really proved that the value of a double integral is equal to the value of the corresponding 2 single integrals in either order of integration, but provided the function is continuous, this is true. The result is chosen Fubini'south Theorem, which we state here without proof.

Theorem 4.xv. Fubini'southward Theorem for Rectangular Regions of Integration.

Let \(f(ten,y)\) be a continuous part defined over the rectangle \(R=[a,b]\times[c,d]\text{.}\) Then the society of integration does not matter, and then

\begin{equation*} \iint_R f(x,y) \, dA = \int_c^d\int_a^b f(x,y)\,dx\,dy = \int_a^b\int_c^d f(10,y)\,dy\,dx\text{.} \end{equation*}

We provide ane example, where we compute the book under a surface in ii ways by switching the order of integration.

Example 4.sixteen. Compute Book in Two Means.

Nosotros compute \(\ds\iint_R 1+(x-1)^2+4y^2\,dA\text{,}\) where \(R=[0,three]\times[0,2]\text{,}\) in two ways.

Solution

First,

\brainstorm{align*} \int_0^3\int_0^2 1+(x-one)^2+4y^2\,dy\,dx \amp =\int_0^3\left. y+(x-1)^2y+{iv\over 3}y^three\correct|_0^2\,dx\\ \amp =\int_0^3 2+2(x-1)^2+{32\over iii}\,dx\\ \amp =\left. 2x + {2\over 3}(x-1)^3+{32\over 3}x\right|_0^3\\ \amp =6+{ii\over 3}\cdot 8 + {32\over iii}\cdot3-(0-1\cdot{2\over3}+0)\\ \amp =44\text{.} \cease{align*}

In the other order:

\begin{align*} \int_0^2\int_0^3 1+(ten-ane)^2+4y^two\,dx\,dy \amp =\int_0^2\left. ten+{(x-1)^iii\over3}+4y^2x\correct|_0^three\,dy\\ \amp =\int_0^2 3+{8\over3}+12y^ii+{1\over3}\,dy\\ \amp =\left. 3y+{viii\over3}y+4y^3+{1\over3}y\right|_0^2\\ \amp =six+{xvi\over3}+32+{ii\over3}\\ \amp =44\text{.} \end{marshal*}

Note: In this example there is no particular reason to favor one direction over the other; in some cases, i direction might be much easier than the other, so it's usually worth considering the two different possibilities.

Subsection 4.2.3 Computing Double Integrals over whatever Region

Frequently we will be interested in a region that is not simply a rectangle. Permit's compute the volume nether the surface \(x+2y^2\) higher up the region described by \(0\le 10\le1\) and \(0\le y\le x^2\text{,}\) shown below.

In principle in that location is zero more difficult nearly this problem. If we imagine the three-dimensional region nether the surface and in a higher place the parabolic region every bit an oddly shaped loaf of bread, we can however piece information technology up, approximate the volume of each slice, and add these volumes up. For example, if we slice perpendicular to the \(x\)-axis at \(x_i\text{,}\) the thickness of a slice will exist \(\Delta x\) and the area of the slice volition exist

\begin{equation*} \int_0^{x_i^2} x_i+2y^2\,dy\text{.} \end{equation*}

When nosotros add these up and accept the limit every bit \(\Delta x\) goes to 0, nosotros get the double integral

\begin{marshal*} \int_0^1 \int_0^{x^2} ten+2y^2\,dy\,dx \amp =\int_0^i \left.xy+{2\over3}y^3\right|_0^{x^2}\,dx\\ \amp =\int_0^1 x^3+{two\over3}ten^vi\,dx\\ \amp =\left. {x^iv\over4}+{2\over21}x^7\right|_0^i\\ \amp ={i\over4}+{two\over21}={29\over84}\text{.} \end{align*}

We could just as well slice the solid perpendicular to the \(y\)-axis, in which case nosotros get

\begin{align*} \int_0^1 \int_{\sqrt y}^1 x+2y^2\,dx\,dy \amp =\int_0^one \left.{x^2\over2}+2y^2x\right|_{\sqrt y}^1 \,dy\\ \amp =\int_0^one {1\over2}+2y^2-{y\over2}-2y^ii\sqrt y\,dy\\ \amp =\left.{y\over2}+{2\over3}y^3-{y^2\over4}-{iv\over7}y^{7/2}\right|_0^1\\ \amp ={i\over2}+{ii\over3}-{1\over4}-{4\over7}={29\over84}\text{.} \end{align*}

What is the average height of the surface over this region? As before, it is the volume divided past the surface area of the base, but at present we need to utilise integration to compute the area of the base, since information technology is not a simple rectangle. The area is

\brainstorm{equation*} \int_0^1 x^ii\,dx={ane\over3}\text{,} \terminate{equation*}

so the average height is

\begin{equation*} \frac{29}{84}\div\frac{1}{iii} = \frac{29}{28}\text{.} \finish{equation*}

Although we accept not proven that the order of integration can exist switched, we nevertheless capture our results and state the general version of Fubini'due south Theorem without proof.

Theorem 4.17. Fubini'south Theorem for General Regions of Integration.

Let \(f(10,y)\) be a continuous function defined over the region \(R\text{.}\)

1. If \(R\) is bounded by \(a \leq x \leq b,\ g_1(x) \leq y \leq g_2(10)\text{,}\) with \(g_1\) and \(g_2\) continuous on \([a,b]\text{,}\) then

   \brainstorm{equation*} \iint_R f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(10)} f(x,y)\, dy\,dx\text{.} \end{equation*}

2. If \(R\) is bounded by \(c \leq y \leq d, \ h_1(y) \leq x \leq h_2(y)\text{,}\) with \(h_1\) and \(h_2\) continuous on \([c,d]\text{,}\) so

   \brainstorm{equation*} \iint_R f(x,y)\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\, dx \,dy\text{.} \end{equation*}

Notation: Although Fubini'southward Theorem tells the states that the guild of integration does not matter in a double integral, the theorem does not tell u.s.a. which of the double integrals is easier to compute. Experience through do allows united states to decide whether to choose to fix a double integral with \(dx\,dy\) or \(dy\,dx\text{.}\)

We explore the order of integration with one more instance of a double integral.

Instance 4.18. Volume of Region.

Find the volume under the surface \(\ds z=\sqrt{1-10^2}\) and above the triangle formed by \(y=ten\text{,}\) \(x=ane\text{,}\) and the \(x\)-axis.

Solution

Let's consider the two possible means to set this up:

\brainstorm{equation*} \int_0^i \int_0^x \sqrt{i-x^two}\,dy\,dx \qquad\hbox{or}\qquad \int_0^1 \int_y^1 \sqrt{1-10^2}\,dx\,dy\text{.} \finish{equation*}

Which appears easier? In the offset, the first (inner) integral is like shooting fish in a barrel, considering we need an antiderivative with respect to \(y\text{,}\) and the entire integrand \(\ds\sqrt{1-x^two}\) is constant with respect to \(y\text{.}\) Of course, the second integral may be more than difficult. In the second, the first integral is mildly unpleasant—a trig substitution. So let's try the offset i, since the first step is like shooting fish in a barrel, and run across where that leaves us.

\begin{equation*} \int_0^1 \int_0^x \sqrt{ane-x^ii}\,dy\,dx= \int_0^1 \left. y\sqrt{1-ten^ii}\right|_0^x\,dx= \int_0^1 10\sqrt{one-x^ii}\,dx\text{.} \stop{equation*}

This is quite easy, since the commutation \(u=i-x^2\) works:

\brainstorm{equation*} \int x\sqrt{1-x^2}\,dx=-{1\over 2}\int \sqrt u\,du ={i\over3}u^{3/two}=-{1\over3}(i-x^2)^{iii/2}\text{.} \end{equation*}

And then

\begin{equation*} \int_0^1 x\sqrt{1-10^2}\,dx=\left. -{1\over3}(i-x^2)^{3/2}\correct|_0^1 ={ane\over3}\text{.} \stop{equation*}

This is a skillful example of how the order of integration tin impact the complexity of the problem. In this example it is possible to do the other order, only it is a flake messier. In some cases one order may lead to an integral that is either difficult or impossible to evaluate; it'southward usually worth considering both possibilities before going very far.

Exercises for Section four.2.

Exercise 4.2.one.

Compute the post-obit double integrals.

1. \(\ds \int_{0}^{2}\int_{0}^{iv} 1+x \,dy\,dx\)

   Respond Solution

   \begin{equation*} \brainstorm{split} \int_0^ii \int_0^iv (one+x)\,dy\,dx \amp = \int_0^2 (ane+x) y\big\vert_0^4\,dx \\ \amp= \int_0^2 4(i+x)\,dx = 4\left[x+\frac{x^2}{2}\right]_0^2 = 16.\end{dissever} \end{equation*}

2. \(\ds \int_{-one}^{1}\int_{0}^{2} x+y\,dy\,dx\)

   Answer Solution

   \begin{equation*} \brainstorm{split up} \int_{-ane}^ane\int_0^2 (x+y)\,dy\,dx \amp= \int_{-one}^1 \left[xy + \frac{y^ii}{2}\right]_0^ii \\ \amp= \int_{-i}^2 \left(2x+ii\right)\,dx = \left[x^two+2x\right]_{-1}^1 = 4. \end{split up} \end{equation*}

3. \(\ds \int_{1}^{2}\int_{0}^{y} xy \,dx\,dy\)

   Respond Solution

   \(\begin{aligned}\int_1^ii\int_0^y xy \,dx\,dy \amp = \int_1^2 y \int_0^y x \,dx\,dy \\ \amp = \int_1^2 y\left[\frac{1}{ii}10^2\right]_0^y\,dy \\ \amp = \int_1^2 \frac{ane}{two}y^iii\,dy = \frac{one}{2}\left[\frac{1}{4}y^4\right]_1^2 = \frac{fifteen}{8} \end{aligned}\)

4. \(\ds \int_{0}^{1}\int_{y^two/ii}^{\sqrt y} \,dx\,dy\)

   Answer Solution

   \begin{equation*} \begin{split} \int_0^1\int_{y^2/2}^{\sqrt{y}}\,dx\,dy \amp= \int_0^1 10\bigg\vert_{y^2/2}^{\sqrt{y}} \,dy \\ \amp= \int_0^1 \sqrt{y}-\frac{y^2}{ii}\,dy\\ \amp \left[\frac{2y^{3/two}}{3}-\frac{y^3}{six}\right]_0^1 = \frac{1}{2} \end{split} \end{equation*}

5. \(\ds \int_{i}^{two}\int_{one}^{ten} {10^2\over y^2}\,dy\,dx\)

   Reply Solution

   \brainstorm{equation*} \begin{divide} \int_1^2\int_1^x \frac{10^ii}{y^ii}\,dy\,dx \amp= \int_1^2 x^two \left[-\frac{ane}{y}\right]_1^x\,dx \\ \amp= \int_1^two x^two \left(-\frac{1}{x}+1\correct)\,dx = \int_1^two \left(-x + x^2\right)\,dx\\ \amp= \left[-\frac{10^ii}{2} + \frac{x^3}{three}\right]_1^2 = \frac{5}{six} \end{split} \end{equation*}

6. \(\ds \int_{0}^{1}\int_{0}^{ten^two} y due east^x\,dy\,dx\)

   Answer Solution

   \brainstorm{equation*} \begin{split}\int_0^1 \int_0^{x^2} e^ten y\,dy\,dx = \int_0^1 \frac{e^x y^2}{2} \big\vert_0^{ten^two}\,dx = \int_0^ane \frac{e^x ten^iv}{ii}\,dx \end{equation*}

   Nosotros now use integration by parts to solve the respective indefinite integral, with \(u=x^4\text{,}\) \(dv = due east^xdx\text{,}\) and and so \(v=e^ten\text{,}\) and \(du=4x^3\,dx\text{:}\)

   \brainstorm{equation*} \int \frac{due east^x ten^4}{2}\,dx = \frac{1}{2} \left[x^4e^10 - 4 \int due east^x ten^3\,dx \right] \end{equation*}

   We now continue past applying Integration by parts twice more with \(dv = due east^x\,dx\text{:}\)

   \begin{equation*} \brainstorm{split} \int \frac{east^x 10^4}{2}\,dx \amp= \frac{1}{2} \left[x^4e^x - 4 \int e^10 x^3\,dx \correct] \\ \amp= \frac{1}{2} \left[ten^four east^x - 4x^3e^x + 12 \int e^10 x^2\,dx \right]\\ \amp= \frac{one}{2} \left[ten^4 e^x - 4x^3e^x + 12x^2e^10 - 24 \int due east^x x\,dx \right]\\ \amp= \frac{1}{two} \left[x^4 eastward^x - 4x^3e^ten + 12x^2e^ten - 24xe^ten + 24e^x \,dx \right] \end{split} \end{equation*}

   Hence, we have

   \begin{equation*} \begin{split} \int_0^1 \int_0^{10^2} \frac{y}{eastward^10}\,dy\,dx \amp= \int_0^1 \frac{due east^x 10^4}{2}\,dx \\ \amp=\frac{1}{2} \left[x^4 eastward^x - 4x^3e^x + 12x^2e^x - 24xe^10 + 24e^10 \,dx \right]_0^i\\ \amp= \frac{9e}{2} - 12.\stop{split} \stop{equation*}

7. \(\ds \int_{0}^{\sqrt{\pi/2}}\int_{0}^{10^2} x\cos y\,dy\,dx\)

   Answer Solution

   \(\begin{aligned}\int_0^one\int_{y^two/2}^{\sqrt{y}}\,dx\,dy \amp = \int_0^1 x\bigg\vert_{y^2/2}^{\sqrt{y}} \,dy \\ \amp = \int_0^one \sqrt{y}-\frac{y^2}{two}\,dy\\ \amp = \left[\frac{2y^{3/2}}{3}-\frac{y^3}{6}\right]_0^1 = \frac{1}{ii} \stop{aligned}\)

8. \(\ds \int_{0}^{\pi/ii}\int_{0}^{\cos\theta}r^ii(\cos\theta-r) \,dr\,d\theta\)

   Reply Solution

   \(\brainstorm{aligned}\int_0^{\pi/2}\int_0^{\cos\theta} r^2 \left(\cos\theta - r\correct)\,dr\,d\theta \amp =\int_0^{\pi/2} \left[\frac{r^3}{three}\cos\theta - \frac{r^iv}{four}\right]_0^{\cos\theta} \,d\theta \\ \amp =\int_0^{\pi/2} \left[\frac{\cos^4\theta}{3}-\frac{\cos^4\theta}{iv}\right]\,d\theta = \int_0^{\pi/2} \frac{\cos^4\theta}{12} \,d\theta \finish{aligned}\) Now utilize the identity \(\displaystyle{\cos^2\theta = \frac{1}{2}(1+\cos (2\theta))}\text{:}\)

   \begin{equation*} \brainstorm{split} \cos^4\theta \amp = (\cos^2\theta)^2 = \frac{1}{2}(1+\cos 2\theta)^ii \\ \amp = \frac{1}{4} \left(1 + \cos^22\theta + 2 \cos ii\theta\correct) \\ \amp = \frac{1}{four}\left(1 + \frac{one}{2}(ane+\cos 4\theta) + 2 \cos 2\theta\right) \\ \amp = \frac{3}{8} + \frac{1}{8} \cos 4\theta + \frac{ane}{ii}\cos 2\theta. \end{split} \end{equation*}

   Therefore,

   \begin{equation*} \begin{separate} \int_0^{\pi/2} \frac{\cos^4\theta}{12} \,d\theta \amp = \frac{1}{12}\int_0^{\pi/2} \left(\frac{3}{eight} + \frac{one}{viii} \cos 4\theta + \frac{1}{2}\cos 2\theta\right)\\ \amp = \frac{1}{12}\left[\frac{iii\theta}{8} + \frac{\sin 4\theta}{32} + \frac{\sin ii\theta}{4}\right]_0^{\pi/2}\\ \amp = \frac{\pi}{64}. \end{split up} \terminate{equation*}

   We observe that

   \brainstorm{equation*} \int_0^{\pi/2}\int_0^{\cos\theta} r^ii \left(\cos\theta - r\right)\,dr\,d\theta = \frac{\pi}{64} \text{.} \end{equation*}

9. \(\ds \int_0^1\int_{\sqrt{y}}^1 \sqrt{10^3+1}\,dx\,dy\)

   Answer

   \(\frac{2}{9}\left(2\sqrt{2}-1\correct)\)

   Solution

   First, we detect that we cannot evaluate the integral

   \begin{equation*} \int \sqrt{x^three+1} \,dx, \terminate{equation*}

   and and so nosotros try switching the club of integration. The region of integration is equally follows:

   

   Therefore,

   \begin{equation*} \int_0^i \int_{\sqrt{y}}^1 \sqrt{x^3+ane} \,dx \,dy = \int_0^1 \int_0^{10^2} \sqrt{ten^3+ane} \,dy \,dx = \int_0^1 x^2 \sqrt{x^3+ane} \,dx. \finish{equation*}

   We now carry out the integration by making the following substitution: let \(u=x^three+i\text{,}\) then \(du = 3x^two\,dx\text{.}\) And so equally \(ten\) goes from \(0 \to 1\text{,}\) \(u\) goes from \(1 \to 2\text{:}\)

   \begin{equation*} \int_0^1 10^2 \sqrt{x^three+1} \,dx = \int_1^2 \frac{1}{3} \sqrt{u}\,du = \frac{2}{nine} u^{3/2}\big\vert_1^2 = \frac{ii}{9} \left(ii\sqrt{2}-i\right). \finish{equation*}

   Therefore, we have plant that

   \begin{equation*} \int_0^ane \int_{\sqrt{y}}^1 \sqrt{x^3+i} \,dx \,dy = \frac{two}{9}\left(2\sqrt{2}-ane\right). \finish{equation*}

10. \(\ds \int_0^1 \int_{x^two}^1 x\sqrt{one+y^2}\,dy\,dx\)

    Answer Solution

    Beginning, detect that we cannot integrate

    \brainstorm{equation*} \int x\sqrt{1+y^2}\,dy. \stop{equation*}

    Therefore, we switch the society of integation:

    

    Therefore,

    \brainstorm{equation*} \int_0^1 \int_{x^2}^1 10\sqrt{1+y^2} \,dy \,dx = \int_0^1 \int_0^{\sqrt{y}} x\sqrt{1+y^2} \,dx \,dy = \int_0^one \frac{y}{ii} \sqrt{ane+y^2}\,dy. \terminate{equation*}

    Now utilize the substitution: \(u=one+y^2\) with \(du = 2y\,dy\text{.}\) And so as \(y\) goes from \(0 \to one\text{,}\) we have \(u\) going from \(1 \to 2\text{.}\) Hence,

    \begin{equation*} \int_0^1 \frac{y}{2} \sqrt{1+y^ii}\,dy = \int_1^two \frac{\sqrt{u}}{4}\,du = \frac{u^{3/2}}{6}\big\vert_1^two = \frac{i}{half-dozen} \left(2\sqrt{2}-1\right). \cease{equation*}

    Therefore, nosotros have shown that

    \brainstorm{equation*} \int_0^1 \int_{x^2}^1 x\sqrt{1+y^two} \,dy \,dx = \frac{1}{6} \left(ii\sqrt{2}-1\right). \end{equation*}

11. \(\ds \int_0^one \int_0^y {2\over\sqrt{1-ten^2}}\,dx\,dy\)

    Answer Solution

    We will observe that this problem is easier if we first switch the order of integration:

    

    Therefore,

    \begin{equation*} \int_0^i \int_0^y \frac{2}{\sqrt{1-x^2}}\,dx \,dy = \int_0^ane \int_x^i \frac{two}{\sqrt{i-10^ii}}\,dy \,dx = \int_0^ane \frac{ii(1-ten)}{\sqrt{ane-x^ii}}\,dx. \terminate{equation*}

    Hence, we split this integral into two:

    \begin{equation*} \int_0^1 \frac{2}{\sqrt{ane-10^2}}\,dx = two\sin^{-1}(x)\big\vert_0^i = \pi. \terminate{equation*}

    For the 2d integral, we utilise the following commutation: \(u=one-x^ii\) with \(du=-2x\,dx\text{.}\) Therefore,

    \begin{equation*} \int_0^1 \frac{-2x}{\sqrt{1-10^two}}\,dx = \int_0^1 \frac{i}{\sqrt{u}}\,du = -two. \stop{equation*}

    Altogether, we find

    \begin{equation*} \int_0^one \int_0^y \frac{2}{\sqrt{1-x^2}}\,dx \,dy = \pi -2. \terminate{equation*}

12. \(\ds \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy\)

    Answer Solution

    We showtime switch the lodge of integration:

    

    Therefore,

    \brainstorm{equation*} \int_0^ane \int_{3y}^iii e^{x^2}\,dx\,dy = \int_0^iii\int_0^{x/3} eastward^{ten^ii}\,dy\,dy = \int_0^iii \frac{x}{3} e^{x^2}\,dx. \terminate{equation*}

    This is an integral which we tin can solve using substitution: permit \(u=x^2\) with \(du=2x\,dx\text{.}\) So,

    \brainstorm{equation*} \int_0^3 \frac{x}{3} e^{x^ii}\,dx = \int_0^9 \frac{1}{6} e^u\,du = \frac{1}{6} \left(e^9-i\right). \end{equation*}

    We have found that

    \begin{equation*} \int_0^1 \int_{3y}^3 e^{x^2}\,dx\,dy = \frac{1}{half-dozen} \left(east^9-1\right). \cease{equation*}

13. \(\ds \int_{-1}^1\int_0^{1-10^ii} 10^ii-\sqrt{y}\,dy\,dx\)

    Answer

    \(\ds {4\over15}-{\pi\over4}\)

14. \(\ds \int_{0}^{\sqrt2/2}\int_{-\sqrt{1-2x^two}}^{\sqrt{1-2x^2}} ten\,dy\,dx\)

    Answer

Do 4.2.2.

Determine the volume of the region that is bounded as follows:

1. below \(z=1-y\) and to a higher place the region \(-1\le x\le 1\text{,}\) \(0\le y\le ane-10^2\)

   Answer Solution

   The area which bounds the surface in the \(ten\)-\(y\)-plane is shown below.

   

   Therefore, the volume is computed as follows:

   \begin{equation*} \brainstorm{split up} V= \int_{-1}^one \int_{0}^{1-10^two} (1-y)\,dy\,dx \amp= \int_{-1}^1 y-\frac{y^two}{two} \big\vert_0^{1-x^2} \,dx\\ \amp= \frac{ane}{2}\int_{-one}^1 (one-10^four) \,dx = \frac{1}{2}\left[x-\frac{x^five}{v}\right]_{-one}^ane = \frac{4}{5}.\finish{separate} \end{equation*}

2. enclosed by \(y=x^two\text{,}\) \(y=4\text{,}\) \(z=x^2\text{,}\) \(z=0\)

   Answer Solution

   In the \(10\)-\(y\)-plane, the surface is bounded by:

   

   And so we set up the post-obit integral to compute the volume:

   \begin{equation*} \begin{split} V=\int_{-2}^two \int_{x^2}^4 \int_0^{x^2} \,dz\,dy\,dx \amp= \int_{-two}^two \int_{x^ii}^4 x^ii\,dy\,dy \\ \amp= \int_{-2}^2 -x^four+4x^two\,dx = \frac{128}{15}.\finish{divide} \cease{equation*}

3. in the first octant divisional by \(y^2=4-x\) and \(y=2z\)

   Respond Solution

   The region is bounded by the following cross-sections:

   

   Therefore, we compute the following every bit follows:

   \begin{equation*} V = \int_0^4 \int_0^{\sqrt{4-x}} \frac{y}{2} \,dy\,dx=2. \end{equation*}

4. in the first octant bounded past \(y^ii=4x\text{,}\) \(2x+y=4\text{,}\) \(z=y\text{,}\) and \(y=0\)

   Answer Solution

   In the first octant of the \(ten\)-\(y\)-plane, the surface is bounded as follows:

   

   Therefore, we ready the integral:

   \begin{equation*} Five = \int_0^1 \int_{ii\sqrt{x}}^{4-2x} \int_0^y dz\,dy\,dx = \frac{11}{iii}. \end{equation*}

5. in the get-go octant bounded by \(x+y+z=9\text{,}\) \(2x+3y=18\text{,}\) and \(x+3y=9\text{.}\)

   Answer Solution

   In the \(x\)-\(y\)-airplane, the region is divisional by:

   

   Therefore, we compute the volume of the region:

   \begin{equation*} V = \int_0^nine \int_{(ix-x)/iii}^{ii(nine-x)/three} (nine-x-y) \,dy\,dx = \frac{81}{2}. \end{equation*}

6. in the start octant bounded by \(x^2+y^2=a^2\) and \(z=x+y\)

   Answer Solution

   In the \(x\)-\(y\)-plane, the region is bounded equally follows:

   

   Therefore, we ready the following integral to compute the book:

   \begin{equation*} V = \int_{-a}^a \int_{-\sqrt{a^2-10^2}}^{\sqrt{a^2-x^2}} (x+y)\,dy\,dx = \frac{2}{3} a^3. \end{equation*}

7. bounded by \(4x^ii+y^2=4z\) and \(z=ii\)

   Answer Solution

   This book is an elliptic paraboloid. For fixed \(z\text{,}\) the cantankerous-sections of the volume in the \(x\)-\(y\) airplane are ellipses. Therefore, we consider this to be the volume of the surface \(z= \frac{i}{4}\left(4x^two+y^ii\right)\) below the ellipse \(\frac{ten^two}{2} + \frac{y^2}{viii}=1\text{:}\)

   \begin{equation*} V = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{viii-4x^2}}^{\sqrt{8-4x^2}} \,dy\, dx = 4\pi. \end{equation*}

8. bounded by \(z=x^two+y^ii\) and \(z=four\)

   Respond Solution

   The book bounded by the paraboloid \(z=ten^two+y^2\) and the plane \(z=4\) is the volume above the surface \(z=x^2+y^2\) below the circle \(iv=x^two+y^2\text{.}\) This volume tin can exist computed by:

   \begin{equation*} V = \int_{-ii}^2 \int_{-\sqrt{4-y^two}}^{\sqrt{4-y^2}} (10^2+y^2) \,dx \,dy = 8\pi. \end{equation*}

9. nether the surface \(z=xy\) and higher up the triangle with vertices \((i,1,0)\text{,}\) \((4,1,0)\text{,}\) \((i,2,0)\) Answer Solution

   The area which premises the region in the \(x\)-\(y\)-plane is shown below.

   

   Therefore, if we let \(ten\) vary from \(1\) to \(four\text{,}\) we run across that \(y\) is divisional between \(1\) and the line \(y=\frac{7}{iii}-\frac{x}{3}\text{.}\) And then the volume under the surface \(z=xy\) and above the area shown in green is calculated by

   \brainstorm{equation*} \brainstorm{split up} \int_1^4\int_1^{7/3-x/three} xy\,dy\,dx \amp = \int_1^4 x\left[\frac{1}{2}y^2\right]_1^{7/three-x/3} \,dx\\ \amp =\int_1^4 \frac{x}{18}\left(x^2-14x+twoscore\correct)\,dx \\ \amp = \left[\frac{i}{eighteen}\left(\frac{x^4}{4}-\frac{14x^three}{3}+20x^two\correct)\right]_1^4 =\frac{31}{8}. \stop{split up} \end{equation*}

Do 4.two.3.

Evaluate \(\ds\iint x^2\,dA\) over the region in the beginning quadrant bounded by the hyperbola \(xy=sixteen\) and the lines \(y=10\text{,}\) \(y=0\text{,}\) and \(x=8\text{.}\)

Answer Solution

The book is divisional by the post-obit area in the \(10\)-\(y\)-airplane:

Therefore, nosotros compute the following integrals:

\begin{equation*} V = \int_0^4 \int_0^x 10^two \,dy\,dx + \int_4^8 \int_0^{16/x} x^two \,dy \,dx = 448. \end{equation*}

Exercise 4.two.4.

A swimming pool is round with a 40 meter diameter. The depth is constant forth due east-west lines and increases linearly from 2 meters at the south end to 7 meters at the north stop. Find the volume of the puddle.

Respond Solution

The domain of the density function is the circular area shown below:

where \(x\) and \(y\) are measured in m. Permit \(z\) denote the density in metres every bit a function of \(10\) and \(y\text{.}\) The density is a plane with the post-obit cross-sections:

Therefore, the density can be expressed every bit

\brainstorm{equation*} z=f(ten,y) = \frac{10+y}{8}+\frac{ix}{2}. \cease{equation*}

Hence, the volume of the pool can be computed by the post-obit integral:

\begin{equation*} V = \int_{-20}^{20} \int_{-\sqrt{20^2-x^2}}^{\sqrt{twenty^2-x^2}} \left[\frac{x+y}{8}+\frac{nine}{2}\right]\,dy\,dx = 1800\pi \ \text{one thousand}^3. \end{equation*}

Exercise 4.two.5.

Find the boilerplate value of \(f(x,y)=due east^y\sqrt{10+e^y}\) on the rectangle with vertices \((0,0)\text{,}\) \((4,0\)), \((4,1)\) and \((0,ane)\text{.}\)

Answer

\(\ds{(e^2+8e+16)\over15}\sqrt{e+4}-{5\sqrt5\over3}-{e^{five/2}\over15} +{1\over15}\)

Solution

We wish to compute the boilerplate value of $f(x,y) = eastward^y\sqrt{x+e^y}$ over the rectangle shown below:

The total area of the rectangle is

\begin{equation*} \int_0^4\int_0^1 \,dy\,dx = 4, \stop{equation*}

and so we compute

\begin{equation*} f_{avg} = \frac{1}{4} \int_0^4\int_0^1 east^y\sqrt{ten+east^y}\, dy\,dx. \finish{equation*}

Evaluating this integral, nosotros find

\begin{equation*} f_{avg} = \frac{1}{15}\left[1-25\sqrt{5}-e^{5/2}+16\sqrt{4+e} +8e\sqrt{4+e} +e^two\sqrt{4+due east}\right]. \finish{equation*}

Exercise 4.2.6.

Opposite the guild of integration on each of the following integrals

1. \(\ds\int_0^9 \int_0^{\sqrt{ix-y}} f(x,y)\,dx \,dy\)

   Answer

   \(\ds\int_0^3\int_0^{ix-x^2} f(x,y)\,dx\,dy\)

   Solution

   The integral

   \begin{equation*} \int_0^9 \int_0^{\sqrt{nine-y}} f(x,y)\,dx\,dy \end{equation*}

   gives the volume nether the surface \(z=f(x,y)\) and above the region in the \(x\)-\(y\)-airplane shown beneath:

   

   So if instead we permit \(x\) vary from 0 to 3, we run across that \(y\) is bounded betwixt \(y=0\) and the curve \(x=\sqrt{9-y} \implies y = 9-10^2\) (for \(x \in [0,3]\)). Therefore,

   \begin{equation*} \int_0^ix \int_0^{\sqrt{ix-y}} f(10,y)\,dx\,dy = \int_0^iii\int_0^{nine-x^2} f(x,y)\,dx\,dy\text{.} \terminate{equation*}

2. \(\ds\int_1^2 \int_0^{\ln x} f(x,y) \; dy \; dx\)

   Respond

   \(\int_0^{\ln(2)} \int_{e^y}^{2} f(x,y)\,dx\,dy\)

   Solution

   The area of integration is shown below:

   

   Therefore,

   \begin{equation*} \int_1^2 \int_0^{\ln x} f(10,y)\,dy\,dx = \int_0^{\ln(2)} \int_{due east^y}^{2} f(x,y)\,dx\,dy. \terminate{equation*}

3. \(\ds\int_0^1 \int_{\arcsin y}^{\pi/2} f(x,y) \; dx \; dy\)

   Reply

   \(\int_0^{\pi/ii} \int_0^{\sin x} f(x,y) \,dy\,dx\)

   Solution

   The area of integration is shown below:

   

   Therefore, nosotros see that

   \begin{equation*} \int_0^1 \int_{\arcsin(y)}^{\pi/2} f(x,y)\,dx\,dy = \int_0^{\pi/two} \int_0^{\sin x} f(10,y) \,dy\,dx. \end{equation*}

4. \(\ds\int_0^1 \int_{4x}^{4} f(x,y) \; dy \; dx\)

   Answer

   \(\int_0^4 \int_0^{y/4} f(x,y)\,dx \,dy\)

   Solution

   The area of integration is shown below:

   

   Therefore, we run into that

   \begin{equation*} \int_0^ane \int_{4x}^4 f(10,y)\,dy\,dx = \int_0^4 \int_0^{y/4} f(x,y)\,dx \,dy. \stop{equation*}

5. \(\ds\int_0^3 \int_{0}^{\sqrt{9-y^2}} f(10,y) \; dx \; dy\)

   Answer

   \(\int_0^three \int_0^{\sqrt{ix-x^ii}} f(x,y)\,dy\,dx\)

   Solution

   Note that the circle centred at the origin with radius 3 has equation \(10^ii+y^2=ix\text{,}\) and in the outset quadrant can be expressed equally \(10=\sqrt{9-y^2}\) or \(y=\sqrt{9-x^2}\text{.}\) Therefore, the area of integration is:

   

   Therefore,

   \begin{equation*} \int_0^three\int_0^{\sqrt{9-y^ii}} f(ten,y)\,dx\,dy = \int_0^3 \int_0^{\sqrt{9-ten^2}} f(x,y)\,dy\,dx. \end{equation*}

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Source: https://www.sfu.ca/math-coursenotes/Math%20158%20Course%20Notes/sec_VolumeAvgHeight.html

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