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Chapter 12 Integration

“You cannot teach a man anything; you can only help him find it within himself.” - Galileo Galilei

Calculus III continues to parallel Calculus I and here we are at integration. Let's review integration in one variable before we tackle integration in several variables. By now you have been using anti-derivatives (and hence the Fundamental Theorem of Calculus) to compute integrals for so long that it is worth remembering the original definition for the definite integral. If \(f:\re \to \re\) is a function, then we define \(\dsp \int_a^b f(x) \ dx\) as the limit of Riemann sums. If \(f\) is positive, then this is the limit of sums of areas of rectangles. Now we will define our integrals once again as limits of sums, but this time we will take limits of sums of volumes.

First, let's formally restate the definition of the definite integral from Calculus I.

Definition 12.1.

If \([a,b]\) is a closed interval, then a partition P of \([a,b]\) is an ordered sequence \(a=x_{0}\lt x_{1}\lt x_{2}\lt \ldots \lt x_{n-1}\lt x_{n}=b\text{.}\) The norm or mesh of P is \(max\{x_{i}-x_{i-1}:i=1,2,\ldots , n\}\) and is denoted \(\| P \|\text{.}\)

Definition 12.2.

The integral of f over \([a,b]\), denoted by \(\dsp{\int_{a}^b f}\text{,}\) is defined (assuming the limit exists) as

\begin{equation*} \dsp{\int_{a}^b f}=\dsp{\lim_{\| P \| \to 0} \ \sum_{i=1}^{n} f(\hat{x_i})\cdot (x_{i}-x_{i-1})} \end{equation*}

where \(\hat{x_i}\in [x_{i-1}, x_{i}]\) and \(n\) is the number of divisions of the partition \(P\text{.}\) If the limit does not exist, we say that \(f\) is not integrable.

Of course, this is the limit of the sum of the areas of a collection of rectangles where the width of the rectangles tends toward zero as the mesh of the partition tends toward zero. In Calculus III, we do the same except that we must partition both the x and y-axes and we are summing volumes over rectangles rather than summing areas over intervals.

Definition 12.3.

Given any two sets, \(A\) and \(B\text{,}\) the Cartesian Product of \(A\) and \(B\) is the set

\begin{equation*} A \times B = \{ (a,b) : a \in A \mbox{ and } b \in B \}. \end{equation*}

In the case where \(A\) and \(B\) are closed intervals these are simply rectangles in the plane,

\begin{equation*} [a,b] \times [c,d] = \{(x,y) \mid x\in [a,b], y\in [c,d]\ \} \end{equation*}
Definition 12.4.

If R is the rectangle \(R=[a,b]\times [c,d]\text{,}\) then a partition, U, of R is a partition of \([a,b]\text{,}\) \(P=\{a=x_{0}\lt x_{1}\lt \ldots \lt x_{n}=b\}\) along with a partition of \([c,d], Q =\{c=y_{0}\lt y_{1}\lt y_{2}\lt \ldots \lt y_{m}=d\}\text{.}\) The norm or mesh of U is the largest width of any of the rectangles, \([x_{i-1},x_i] \times [y_{j-1},y_j]\) where \(i = 1,2,\dots n\) and \(j=1,2,\dots m\text{.}\) The norm of \(U\) is denoted by \(\|U\|\text{.}\)

Definition 12.5.

If \(f \in C_{R}\) (i.e. f is continuous on the rectangle R), then

\begin{equation*} \int _{R} f \ dA = \dsp{\lim_{\| U \| \to 0} \ \sum_{i=1}^n \sum_{j=1}^m f(\hat{x_i}, \hat{y_j}) (x_{i}-x_{i-1})(y_{j}-y_{j-1})} \end{equation*}

where

\begin{equation*} (\hat{x_i}, \hat{y_j}) \in [x_{i-1}, x_{i}] \times [y_{j-1}, y_{j}]. \end{equation*}

Thankfully, in Calculus III we won't have to compute any such limits because there is a theorem that states:

\begin{equation*} \dsp{\int_{R} f \ dA = \int _{c}^d \int_{a}^b f(x,y)\ dx\ dy} \end{equation*}

which gives us a straightforward way to compute integrals using anti-derivatives by using the Fundamental Theorem of Calculus from Calculus I:

The next problem reminds us that in Calculus I we can integrate with respect to either \(x\) or \(y\) and obtain the same result. These techniques are especially handy in Calculus III.

Problem 12.7.

Compute the area bounded by \(y=x^2\text{,}\) the x-axis, and \(x=2\) two ways. First, use an integral with respect to x, \(\dsp{\int_{\_}^{\_} \_\_\_ \ dx}\text{,}\) then use an integral with respect to y, \(\dsp{\int_{\_}^{\_} \_\_\_ \ dy}\text{.}\) Sketch a picture to help explain your endpoints of integration.

Most of the theorems that you proved in Calculus I have an analog in Calculus III. The following theorem provides a list of properties for double integrals that won't surprise you. We won't state them again, but they also hold for any integrable function \(f: \re^n \to \re\) where \(n > 2.\)

When we are integrating over a two dimensional region R, we will write \(\int_R f \ dA\text{,}\) where the \(A\) reminds us that we are integrating over a region with area. This will be a double integral. When we are integrating over a three dimensional region, we will write \(\int_R f \ dV\text{,}\) where the \(V\) reminds us that we are integrating over a region with volume. This will be a triple integral.

Problem 12.9.

Compute \(\dsp{\int_{1}^2 \int_{0}^3 2x+3y \ dy\ dx}\) and \(\dsp{\int_{0}^3 \int_{1}^2 2x+3y \ dx\ dy}\text{.}\) This is called “reversing the order of integration.” Sketch the solid that you found the volume of.

Problem 12.10.

Compute and compare these three integrals:

  1. \(\displaystyle \dsp{\int_{1}^2 \int_{x-2}^{0} 2x+3y \ dy\ dx}\)

  2. \(\displaystyle \dsp{\int_{x-2}^{0} \int_{1}^2 2x+3y \ dx\ dy}\)

  3. \(\displaystyle \dsp{\int_{-1}^{0} \int_{1}^{y+2} 2x+3y \ dx\ dy}\)

As you can see from the previous two problems, when the limits of integration contain variables care must be taken in reversing the order of integration. When the limits of integration are numbers, we can reverse the order of the integrals and obtain the same result. This is known as Fubini's Theorem.

Problem 12.12.

Evaluate \(\dsp \int_{B} 4x \ln(y)z \; dV\) over the box \(B= [0, 2] \times [1, 4] \times [-2, 5]\text{.}\) Now change the order of integration and verify the result. How many choices are there for orders of integration?

Problem 12.13.

Let \(f(x,y)=4-x^2-y^2\) over the region \(R=[0,1] \times [0,1].\) Compute \(\dsp \int_0^1 \int_0^1 f(x,y) \ dy \ dx.\) Sketch the solid that this integral represents the volume of.

Problem 12.14.

Let \(f(x,y)=4-x^2-y^2.\) Write an integral for the volume of the solid that that is bounded by \(f\) and by the planes \(z=0, x=0, x=2, y=0,\) and \(y=2.\) Compute this integral.

Problem 12.15.

Compute \(\dsp{\int_{0}^1 \int_{0}^1 \frac{y}{(xy+1)^2}\ \ dx\ dy}.\)

Problem 12.16.

Compute \(\dsp{\int_{0}^{\ln(3)} \int_{0}^1 xye^{xy^2}\ \ dy\ dx}.\)

Problem 12.17.

Compute \(\dsp{\int_{R}\sin(x+y)\ \ dA}\) where \(R=\left[0,\dsp{\frac{\pi}{2}}\right]\times \left[0,\dsp{\frac{\pi}{2}}\right].\)

Problem 12.18.

Compute \(\dsp{\int_{R} xy\sqrt{1+x^2}\ \ dA}\) where \(R=[0,\sqrt{3}]\times [1,2].\)

Thus far we have integrated primarily over domains that were rectangles, but we can also integrate over more general domains.

Problem 12.19.

Let \(f(x,y) = 4x + 2y.\)

  1. Sketch the region in the \(xy\)-plane bounded by by \(x=2\text{,}\) \(x=4\text{,}\) \(y=-x\text{,}\) \(y=x^2\text{.}\)

  2. Compute the volume of the solid below \(f\) and above this region.

Problem 12.20.

Evaluate \(\dsp{\int_{S} xy\ \ dA}\) where S is the region bounded by \(y=x^2\) and \(y=1\text{.}\)

Problem 12.21.

Evaluate \(\dsp{\int_{S} \frac{2}{1+x^2}\ \ dA}\) over the region S determined by the triangle with vertices \((0,0)\text{,}\) \((2,2)\text{,}\) and \((0,2)\) in the x-y plane. Sketch the solid that you found the volume of.

Example 12.22.

Find an integral expression for the volume of one-eighth of the sphere of radius 2 in three ways:

\begin{equation*} \int_{\_}^{\_} \int_{\_}^{\_} \_\_\_\_ \ dx dy = \int_{\_}^{\_} \int_{\_}^{\_} \_\_\_\_ \ dy dz = \int_{\_}^{\_} \int_{\_}^{\_} \int_{\_}^{\_} 1 \ dx dy dz \end{equation*}
Problem 12.23.

Fill in the blanks in order to change the order of integration.

\begin{equation*} \dsp{\int_0^1{\int_0^\frac{1-x}{2} \int_0^{1-x-2y}{f(x,y,z)\ dz\ dy\ dx}}}= \int_\_^\_{\int_\_^\_{\int_\_^\_ f(x,y,z) {dx\ dz\ dy}}} \end{equation*}
Problem 12.24.

Sketch and compute the volume of the solid bounded by \(x^2=4y\text{,}\) \(z=0\text{,}\) and \(5y+9z-45=0\text{.}\) Write the integral both as \(\dsp{\int_{\_}^{\_} \int_{\_}^{\_} \_\_\_\ dx\ dy}\) and \(\dsp{\int_{\_}^{\_} \int_{\_}^{\_} \_\_\_\ dy\ dx}\text{.}\)

Problem 12.25.

Fill in the blanks:

\begin{equation*} \dsp{\int_{0}^1 \int_{-y}^y f(x,y)\ dx\ dy}=\dsp{\int_{\_}^{\_} \int_{\_}^{\_} f(x,y)\ dy\ dx} \end{equation*}
\begin{equation*} \dsp{\int_{0}^2 \int_{y^2}^{2y} f(x,y)\ dx\ dy}=\dsp{\int_{\_}^{\_} \int_{\_}^{\_} f(x,y)\ dy\ dx} \end{equation*}
Problem 12.26.

Compute \(\dsp{\int_{0}^1 \int _{-x}^x e^{x+y}\ dy\ dx}\) directly and by reversing the order of integration.

Coordinate Transformations (Change of Variables)

Loosely speaking, a coordinate transformation is a transformation from one coordinate system to another coordinate system and is also called change of variables. A cleverly chosen coordinate transformation can make a difficult integral easy. There are infinitely many, but the three most common are: the conversion from rectangular coordinates to polar coordinates (used in double integrals), from rectangular coordinates to spherical coordinates (used in triple integrals), and from rectangular coordinates to cylindrical coordinates (used in triple integrals).

In Calculus I when you did a trigonometric substitution, you did a change of variable. Written as a theorem, it would look like this.

Work the following problem, using the given substitution, but keeping all the independent variables in tact, i.e. don't toss out the “t” when you replace \(x\) by \(u(t).\)

Problem 12.28.

Apply Theorem 12.27 to compute:

  1. \(\dsp \int_3^4 \frac{1}{x^2 + 9} \ dx\) using the substitution \(u(t) = 3\tan(t)\)

  2. \(\dsp \int_3^4 \frac{1}{\sqrt{x^2 + 9}} \ dx\) using the substitution \(u(t) = 3\tan(t)\)

For transformation of two variables the theorem is similar.

Definition 12.29.

If \(x : {\re}^2 \to \re\) and \(y : {\re}^2 \to \re\) are differentiable functions, then the Jacobian is the function \(J : {\re}^2 \to \re \) given by

\begin{equation*} J(u,v)= det \begin{pmatrix} x_{u}(u,v) \amp x_{v}(u,v)\\ y_{u}(u,v) \amp y_{v}(u,v) \end{pmatrix} = x_{u}(u,v)y_{v}(u,v)-y_{u}(u,v)x_{v}(u,v). \end{equation*}
Example 12.31.

Consider \(\dsp{\iint e^{\frac{y-x}{y+x}} \ dx\ dy}\) over the trapezoidal region with vertices \((0,1), (1,0), (0,2)\) and \((2,0)\text{.}\) Let \(u=y-x\) and \(v=y+x\text{.}\) Sketch the new region, show that the Jacobian is \(J(u,v)= det \begin{pmatrix} -\frac{1}{2} \amp \frac{1}{2}\\ \frac{1}{2} \amp \frac{1}{2} \end{pmatrix} \text{,}\) and compute the integral \(\dsp{\frac{1}{2} \int_1^2 \int_{-v}^{v} e^{\frac{u}{v}} \; du \; dv = \dots = \frac{3}{4}(e-\frac{1}{e})}\)

Problem 12.32.

Integrate \(\dsp \int_0^4 \int_{\frac{y}{2}}^{\frac{y}{2}+1} \frac{2x-y}{2} \; dx \; dy\text{.}\) Now make the change of variable, \(\dsp u = \frac{2x-y}{2}\) and \(\dsp v=\frac{y}{2}\) and integrate the result.

Polar Coordinates Refresher

In polar coordinates, the point in the plane \(P = (x,y)\) may be denoted by \((r,\theta)\) where \(r\) is the signed distance from \((0,0)\) to \(P\text{,}\) and \(\theta\) is the angle between the vector \(\oa{P}\) and the positive x-axis. We restrict \(\theta\) to the interval \([0, 2\pi]\text{.}\) We say \(r\) is the signed distance to allow equations such as \(r = 4\sin(\theta)\) where if \(\dsp \theta = \frac{11\pi}{6}\) then \(r = -2\text{.}\) The corresponding point would be \((\dsp -2, \frac{11\pi}{6})\) which is the same point as \(\dsp (2, \frac{5\pi}{6})\text{.}\) It follows that \(x, y, r,\) and \(\theta\) are related by the equations:

  1. \(\displaystyle x=rcos\theta,\)

  2. \(\displaystyle y=rsin\theta,\)

  3. \(r=\sqrt{x^2+y^2},\) and

  4. \(\theta= arctan\left(\dsp{\frac{y}{x}}\right)\text{.}\)

Problem 12.33.

Sketch each of the following pairs of polar functions on the same graphs.

  1. \(r=5\cos(\theta)\) and \(\theta = 2\pi/3\)

  2. \(r=2\sin(\theta)\) and \(r=2\cos(\theta)\)

  3. \(r=2 + 2\cos(\theta)\) and \(r=1\)

Example 12.34.

Consider \(\dsp{\iint e^{-x^2-y^2} \ dx\ dy}\) over the circle \(x^2+y^2 = 9\text{.}\) Let \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\text{.}\) Sketch the region, determine the bounds of integration in \(r\) and \(\theta\text{,}\) compute the Jacobian, and show that the integral is \(\dsp{\int_0^{2\pi} \int_{0}{3} r e^{-r^2} \; dr \; d\theta}\)

Problem 12.35.

To make a change of variables from rectangular to polar coordinates, we let \(x\) and \(y\) be the functions, \(x(r,\theta)=rcos\theta\) and \(y(r,\theta)=rsin\theta.\) Show that \(J(r,\theta) = r\) using Theorem 12.30 by filling in the missing computation:

\begin{equation*} J(r,\theta)= det \begin{pmatrix}x_r \amp x_{\theta} \\ y_r \amp y_{\theta} \end{pmatrix} = \dots = r. \end{equation*}
Problem 12.36.

To make a change of variables from polar to rectangular coordinates, we let \(r\) and \(\theta\) be the functions, \(r(x,y)=\sqrt{x^2+y^2}\) and \(\theta(x,y)=arctan(y/x).\) Compute and simplify \(J(x,y)\text{.}\)

Problem 12.37.

Compute \(\dsp{\int_0^{\frac{\pi}{2}}\int_0^{\cos(\theta)}r^2\sin(\theta)\ dr\ d\theta}\text{.}\)

Problem 12.38.

Convert the previous integral to rectangular coordinates and recompute to verify your answer.

Using the previous problem and Theorem 12.30, we now see that when making a change of variable from rectangular to polar coordinates we have,

\begin{equation*} \dsp{\int_{B} f(x,y)\ dx\ dy=\int_{B'} f(r\cos(\theta),r\sin(\theta))\ r\ dr\ d\theta} \end{equation*}

where \(B'\) is the region \(B\) represented in polar coordinates. The next problem demonstrates an intuitive geometric argument supporting this result.

Problem 12.39.

Recall that the area of the sector of a circle with radius \(r\) spanning \(\theta\) radians is \(A=\frac{1}{2} r^2\theta\text{.}\) Let \(0 \lt r_1 \lt r_2\) and \(0 \lt \theta_1 \lt \theta_2 \lt \frac{\pi}{2}.\) Sketch the region in the first quadrant bounded by the two circles \(r=r_1,\) \(r=r_2,\) and the two lines \(\theta = \theta_1,\) and \(\theta = \theta_2.\) Show that the area of the bounded region is \(\dsp{\left(\frac{r_{1}+r_{2}}{2}\right)} (r_{2}-r_{1})(\theta_{2}-\theta_{1})\text{.}\)

Example 12.40.

Write an integral with respect to \(\theta\) that represents the area inside the graph of \(r=\sin(2\theta)\) using limits and demonstrate why \(\dsp{\left(\frac{r_{i}+r_{i+1}}{2}\right)} (r_{i+1}-r_{i})(\theta_{i+1}-\theta_{i})\) appears in the sum. Now write the same area as a double integral with respect to \(r\) and \(\theta\text{.}\) Compute the volume of the cylinder of height 4 above \(r=\cos(2\theta).\) Hopefully, the theme of looking at the areas or volumes of small pieces comes through.

Definition 12.41.

Two circles are concentric if they share a common center, yet have distinct radii.

Definition 12.42.

An annulus is a region in the plane trapped between two concentric circles.

Problem 12.43.

Let \(r_1\) and \(r_2\) be real numbers with \(0 \lt r_1 \lt r_2.\) Let \(R\) be the portion of the annulus centered at the origin, between the two circles of radii \(r_1\) and \(r_2\text{,}\) and above the x-axis. Convert \(\dsp{\int_R{e^{x^2+y^2}\ dA}}\) over the region \(R\) to polar coordinates and compute. Write down the endpoints of integration in rectangular coordinates.

Problem 12.44.

Convert to polar and compute \(\dsp{\int_0^1\int_0^{\sqrt{1-y^2}}\sin(x^2+y^2)\ dx\ dy}\text{.}\)

Problem 12.45.

Sketch the regions bounded by \(r=2\) and \(r=2(1+\cos(\theta))\) from \(\theta=0\) to \(\theta = 2\pi\text{.}\) Compute \(\dsp{\int_R{y\ dA}}\) via polar coordinates over the smallest of the regions.

Problem 12.46.

Sketch \(\dsp{\theta=\frac{\pi}{6}}\) and \(r=4\sin\theta\text{.}\) Compute the area of the smaller of the two regions bounded by the curves in two ways:

  1. First compute \(\dsp{\iint_R{r\ dr\ d\theta}}\text{.}\)

  2. Check your answer by using the formula for the area inside a polar graph, \(\frac{1}{2}\int_\alpha^\beta (f(\theta))^2 \ d\theta\text{.}\)

Problem 12.47.

Convert to polar and compute \(\dsp{\int_1^2\int_0^{\sqrt{2x-x^2}}(x^2+y^2)^{-\frac{1}{2}}\ dy\ dx}\text{.}\)

Problem 12.48.

Find the volume of the solid under \(z=3xy,\) above \(z=0,\) and within \(x^2+y^2=2x\text{.}\)

Problem 12.49.

Find the volume of the solid in the \(1^{st}\) octant (i.e. where \(x>0, y>0, z>0\)) under \(z=x^2+y^2\) and inside the surface \(x^2+y^2=9\text{.}\)

Cylindrical and Spherical Coordinates

You already know how to represent points in the plane in two different ways, rectangular and polar coordinate systems. We wish to represent points in three-space in two new ways referred to as cylindrical and spherical coordinates.

Definition 12.50.

The Cylindrical Coordinate Representation for a point \(P = (x,y,z)\) is denoted by \((r,\ \theta,\ z)\) where \((r,\ \theta)\) are the polar coordinates of the point \((x,y)\) in the x-y-plane and \(z\) remains unchanged (i.e. \(z\) is the height of the point above or below the x-y-plane).

Definition 12.51.

The Spherical Coordinate Representation for a point \(P = (x,y,z)\) is denoted by \((\rho,\ \phi,\ \theta)\) where \(\rho\) is the distance from the point to the origin, \(\phi\) is the angle between \(\oa P\) and positive z-axis, and \(\theta\) is the angle between the x-axis and the projection of \(\oa P\) onto the x-y-plane. We restrict \(\rho \geq 0\text{,}\) \(0 \leq \phi \leq \pi\text{,}\) and \(0 \leq \theta \leq 2\pi\text{.}\)

Example 12.52.

Write an integral using spherical coordinates that represents the volume of an ice cream cone — that is, the volume inside the cone \(\phi = \pi/6\) and inside the sphere \(\rho=6\text{.}\) Of course, we must insert the Jacobian \(\rho^2 \sin(\phi)\text{.}\)

Problem 12.53.

Sketch and write a sentence describing each of the following regions in spherical coordinates.

  1. the region between \(\rho = 2\) and \(\rho = 3\)

  2. the region between \(\phi = \pi/2\) and \(\pi/4\)

  3. the region below \(\rho = 10\) and above \(\phi = \pi/3\)

Example 12.54.

Convert

\begin{equation*} \int_0^2 \int_0^{\sqrt{4-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \sqrt{4-x^2-y^2} \ dz \ dy \ dx \end{equation*}

to both spherical and cylindrical coordinates.

From Problem 12.36, we know that the Jacobian for polar coordinates is

\begin{equation*} J(r,\theta) = det \begin{pmatrix}\cos(\theta) \amp -r \ \sin(\theta) \cr \sin(\theta) \amp r \ \cos(\theta) \end{pmatrix} =r. \end{equation*}

Two of the next three problems generate the Jacobian for cylindrical and spherical coordinate systems. Warning. If you haven't had linear algebra yet, the next two problems require computing the determinant of \(3 \times 3\) matrices. You might skip them, ask me how to compute the determinant of a \(3 \times 3\) matrix, or Google how to do it.

Problem 12.55.

To make a change of variables from rectangular to cylindrical coordinates, we let \(x=x(r,\theta)=r\ \cos\theta\text{,}\) \(y=y(r,\theta)=r\ \sin\theta\text{,}\) and \(z=z\text{.}\) Show that \(J(r,\theta,z) = r\) using Theorem 12.30 by filling in the missing computation:

\begin{equation*} J(r,\theta,z)= det \begin{pmatrix}x_r \amp x_{\theta} \amp x_z\cr y_r \amp y_{\theta} \amp y_z\cr z_r \amp z_{\theta} \amp z_z \end{pmatrix} = \dots = r. \end{equation*}
Problem 12.56.

Consider the solid trapped inside the cylinder \(x^2+y^2=9\text{,}\) above the function \(f(x,y) = 4-x^2-y^2\) and below the plane \(z=10\text{.}\)

  1. Write a triple integral in rectangular coordinates for the volume, but don't compute.

  2. Write a triple integral in cylindrical coordinates for the volume and compute it.

  3. Write the volume as the volume of the cylinder minus the volume under the upside down paraboloid.

Problem 12.57.

To make a change of variables from rectangular to spherical coordinates, we let \(x(\rho,\phi,\theta)=\rho\ \sin\phi\ \cos\theta\text{,}\) \(y(\rho,\phi,\theta)=\rho\ \sin\phi\ \sin\theta\text{,}\) and \(z(\rho,\phi,\theta)=\rho\ \cos\phi.\) Show that \(J(\rho, \phi, \theta) = \rho^2\sin(\phi)\) using Theorem 12.30 by filling in the missing computation:

\begin{equation*} J(\rho, \phi, \theta) = det \begin{pmatrix}x_{\rho} \amp x_{\phi} \amp x_{\theta} \cr y_{\rho} \amp y_{\phi} \amp y_{\theta} \cr z_{\rho} \amp z_{\phi} \amp z_{\theta} \end{pmatrix} = \dots = \rho^2\sin(\phi). \end{equation*}

Summarizing, when integrating with respect to polar, cylindrical, or spherical coordinates, we will always use the appropriate Jacobian as illustrated below.

  1. Polar:

    \begin{equation*} \dsp{\iint \ldots\ r\ dr\ d\theta}, \ \ \ \mbox{where} \ \ x=r \cos(\theta), y = r \sin(\theta) \end{equation*}

  2. Cylindrical:

    \begin{equation*} \dsp{\iiint \ldots\ r\ dr\ d\theta \ dz}, \ \ \ \mbox{where} \ \ x=r \cos(\theta), y = r \sin(\theta), \ \ z = z \end{equation*}

  3. Spherical:

    \begin{equation*} \dsp{\iiint \ldots\ \rho^2\sin(\phi)\ d\rho \ d\phi \ d\theta}, \ \ \ \mbox{where} \ \ x=\rho \sin(\phi) \cos(\theta), y=\rho \sin(\phi) \sin(\theta), \ \ z=\rho \cos(\phi) \end{equation*}

The order of integration could change and you will find times when the choice of the order of integration transforms an apparently hard problem into an easy one.

Problem 12.58.

Compute the volume of the region bounded by the two spheres \(\rho=4\) and \(\rho=6\) using spherical coordinates. Verify using the formula for the volume of a sphere, \(\dsp V = \frac{4\pi}{3}r^3.\)

Problem 12.59.

Sketch and find the volume of the solid outside the cone \(\dsp{\phi=\frac{\pi}{3}}\text{,}\) inside the sphere \(\rho=6\text{,}\) and above the plane \(z=0.\)

Problem 12.60.

Convert to spherical coordinates then evaluate: \(\dsp {\int_{-3}^{3} {\int_{-\sqrt{9-x^{2}}}^{\sqrt{9-x^{2}}} {\int_{-\sqrt{9-x^{2}-z^{2}}}^{\sqrt{9-x^{2}-z^{2}}} \ \ {(x^2+y^2+z^2)^{\frac{3}{2}}\ dy\ dz\ dx}}}}\)

Problem 12.61.

Convert to spherical coordinates and compute: \(\dsp \int_R e^{ (x^2+y^2+z^2)^\frac{3}{2} } \ dV\) where \(R\) is the region \(x^2+y^2+z^2 \leq 1.\)

Problem 12.62.

Find the volume of the given ellipsoids by first writing a triple integral for the volume in rectangular coordinates and then using the recommended change of variables to convert the triple integral into the new coordinate system. If all goes well, the converted integral will be easier to compute because you are transforming the ellipsoid into a sphere.

  1. \(\dsp{\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{36}\leq1}\) via the change of variables \(x=2a\text{,}\) \(y=3b\text{,}\) \(z=6c\)

  2. \(\dsp{\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\leq1}\) via the change of variables \(x=ua\text{,}\) \(y=vb\text{,}\) \(z=wc\)

An Integration Application

The mass of an object can be found by integrating the density function, \(\delta,\) over the entire object. If \(\delta\) is the density function, then the center of mass is given by \((\oa{x},\ \oa{y},\ \oa{z}) = \dsp{\left(\frac{M_{yz}}{m},\ \frac{M_{xz}}{m},\ \frac{M_{xy}}{m} \right)}\) where

  1. \(\displaystyle m = mass = \int \delta(x,y,z) \ dx \ dy \ dz,\)

  2. \(\displaystyle M_{xy}=\iiint {z\ \delta(x,y,z)}\ dx \ dy \ dz,\)

  3. \(M_{xz}=\iiint{y\ \delta(x,y,z)}\ dx \ dy \ dz,\) and

  4. \(\displaystyle M_{yz}=\iiint{x\ \delta(x,y,z)}\ dx \ dy \ dz.\)

The numbers, \(M_{xy}, M_{xz},\) and \(M_{yz}\) are called the moments of the object with respect to \(z\text{,}\) \(y\text{,}\) and \(x\) respectively.

Problem 12.63.

Find the center of the mass of the solid inside \(x^2+y^2=4\text{,}\) outside \(x^2+y^2=1\text{,}\) below \(z=12-x^2-y^2\text{,}\) and above \(z=0\text{,}\) assuming the constant density function \(\rho(x,y,z) = k\text{.}\) Graph the solid and mark the center of mass to see if your answer makes sense!

Problem 12.64.

Find the mass of the solid bounded by \(\dsp z=2-\frac{1}{2}x^{2},\ z=0,\ y=x\) and \(y=0,\) assuming the density is \(\delta(x,y,z)=kz\) where \(k>0.\) Challenge: Write this integral with the various orders of integration, \(dz \ dy \ dx\text{,}\) \(dy \ dx \ dz\text{,}\) \(dz \ dx \ dy\text{,}\) and \(dx \ dz \ dy\text{.}\)

Problem 12.65.

Find the center of the mass of the solid from the previous problem.