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 partial integrals double integrals double integrals - area and volume triple integrals

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17calculus > exam list > calc3 exam A3

This is the third exam for third semester (multi-variable) calculus. This exam contains 8 questions on partial integrals including double and triple integrals and their applications.

### Practice Exam Tips

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
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- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
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IMPORTANT -
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

Exam Details

Time

2 hours

Questions

8

Total Points

85

Tools

Calculator

no

Formula Sheet(s)

1 page, 8.5x11 or A4

Other Tools

none

Instructions:
- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
- Correct notation counts (i.e. points will be taken off for incorrect notation).

(5 points) Evaluate the integral $$\displaystyle{ \int_{1}^{2}{ \int_{0}^{\ln(x)}{ x^3 e^y ~dy } ~dx } }$$.

Problem Statement

(5 points) Evaluate the integral $$\displaystyle{ \int_{1}^{2}{ \int_{0}^{\ln(x)}{ x^3 e^y ~dy } ~dx } }$$.

$$49/20$$

Problem Statement

(5 points) Evaluate the integral $$\displaystyle{ \int_{1}^{2}{ \int_{0}^{\ln(x)}{ x^3 e^y ~dy } ~dx } }$$.

Solution

[ Inside Integral ]

$$\int_{0}^{\ln(x)}{x^3 e^y~dy}$$

$$\left. x^3 e^y \right|_{y=0}^{y=\ln(x)}$$

$$x^3 \left[ e^{\ln(x)} - e^0 \right]$$

$$x^3 \left[ x - 1 \right] = x^4 - x^3$$

[ Outside Integral ]

$$\int_{1}^{2}{x^4 - x^3~dx}$$

$$\displaystyle{ \left[ \frac{x^5}{5} - \frac{x^4}{4} \right]_{1}^{2} }$$

$$[2^5/5 - 2^4/4] - [1/5 - 1/4] = 49/20$$

$$49/20$$

(10 points) Sketch the region of integration and set up (but do not evaluate) the integral $$\displaystyle{ \iint\limits_R{ y^2 ~dA } }$$ where R is the region bounded by $$y=2, y=2-x$$ and $$y=x-2$$.

Problem Statement

(10 points) Sketch the region of integration and set up (but do not evaluate) the integral $$\displaystyle{ \iint\limits_R{ y^2 ~dA } }$$ where R is the region bounded by $$y=2, y=2-x$$ and $$y=x-2$$.

Solution

The region is shown in the plot above. It is better to describe the region horizontally since this allows us to set up only one integral. In this case, $$0 \leq y \leq 2$$ and $$2-y \leq x \leq y+2$$. So our integral is
$$\displaystyle{\int_{0}^{2}{\int_{2-y}^{y+2}{ y^2 ~dx} ~dy}}$$

(10 points) Sketch the region of integration and set up (but do not evaluate) this integral in polar coordinates $$\displaystyle{ \int_{-5}^{5}{ \int_{0}^{\sqrt{25-y^2}}{ 16-x^2-y^2 ~dx }~dy } }$$

Problem Statement

(10 points) Sketch the region of integration and set up (but do not evaluate) this integral in polar coordinates $$\displaystyle{ \int_{-5}^{5}{ \int_{0}^{\sqrt{25-y^2}}{ 16-x^2-y^2 ~dx }~dy } }$$

Solution

The region is shown in the plot above. From the integral, we can determine the ranges on x and y as $$-5 \leq y \leq 5$$ and $$0 \leq x \leq \sqrt{25-y^2}$$. From the last inequality for x, we can determine that we are working with a circle of radius 5. The range on y may give us the impression that we need the entire circle. However, since x starts at zero, this tells us that we have only the right half of the circle.
The standard equation for polar coordinates is $$r^2=x^2+y^2$$. In our case, $$r=5$$ and so our range on r is $$0\leq r\leq 5$$. Since we want only the right half of the circle, our range on the angle is $$-\pi/2 \leq \theta \leq \pi/2$$. Now we can set up our integral.
$$\displaystyle{\int_{-\pi/2}^{\pi/2}{\int_{0}^{5}{ (16-r^2)~rdr}~d\theta}}$$
Since both sets of limits of integration involve real numbers (no variables), the order of integration can be switched with no change to the integrand, i.e. $$\displaystyle{\int_{0}^{5}{\int_{-\pi/2}^{\pi/2}{(16-r^2)~rd\theta}~dr}}$$ is also a correct answer.

(5 points) For the integral in the previous question in rectangular coordinates, set up (but do not evaluate) an integral (also in rectangular coordinates) by switching the order of integration.

Problem Statement

(5 points) For the integral in the previous question in rectangular coordinates, set up (but do not evaluate) an integral (also in rectangular coordinates) by switching the order of integration.

Solution

Looking at the plot, we can see that in the previous question integration was done horizontally (since y ranges from -5 to 5). So we need to integrate vertically. This means that the range on x is $$0 \leq x \leq 5$$ and y ranges from $$-\sqrt{25-x^2}$$ to $$\sqrt{25-x^2}$$. The integrand does not change, so our answer is $$\displaystyle{ \int_{0}^{5}{ \int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}{ 16-x^2-y^2 ~ dy} ~dx } }$$

(10 points) Evaluate $$\displaystyle{ \int_{0}^{\pi}{ \int_{0}^{y}{ \int_{0}^{\sin(x)}{ dz } ~dx } ~dy } }$$

Problem Statement

(10 points) Evaluate $$\displaystyle{ \int_{0}^{\pi}{ \int_{0}^{y}{ \int_{0}^{\sin(x)}{ dz } ~dx } ~dy } }$$

$$\pi$$

Problem Statement

(10 points) Evaluate $$\displaystyle{ \int_{0}^{\pi}{ \int_{0}^{y}{ \int_{0}^{\sin(x)}{ dz } ~dx } ~dy } }$$

Solution

[ Inside Integral ]
$$\displaystyle{ \int_{0}^{\sin(x)}{ dz } = \sin(x) }$$

[ Middle Integral ]
$$\displaystyle{\int_{0}^{y}{\sin(x)~dx} = \left. -\cos(x) \right|_{0}^{y} = -\cos(y)+\cos(0) = 1-\cos(y)}$$

[ Outside Integral ]
$$\displaystyle{\int_{0}^{\pi}{1-\cos(y)~dy} = \left[ y-\sin(y) \right]_{0}^{\pi} = \pi}$$

$$\pi$$

(15 points) Set up (but do not evaluate) an integral to calculate the volume of the prism in the first octant bounded by the planes $$y=3-3x$$ and $$z=2$$.

Problem Statement

(15 points) Set up (but do not evaluate) an integral to calculate the volume of the prism in the first octant bounded by the planes $$y=3-3x$$ and $$z=2$$.

Solution

From the equations in the problem statement, we know that the range on z is $$0 \leq z \leq 2$$. The ranges on x and y can be derived from the above graph. There are two possible valid orientations, vertical and horizontal.

 vertical orientation

In this case the ranges on x and y are $$0 \leq x \leq 1$$ and $$0 \leq y \leq 3-3x$$. This gives these three possible correct answers.

 $$\displaystyle{\int_{0}^{2}{\int_{0}^{1}{\int_{0}^{3-3x}{1~dy}~dx}~dz}}$$ $$\displaystyle{\int_{0}^{1}{\int_{0}^{2}{\int_{0}^{3-3x}{1~dy}~dz}~dx}}$$ $$\displaystyle{\int_{0}^{1}{\int_{0}^{3-3x}{\int_{0}^{2}{1~dz}~dy}~dx}}$$
 horizontal orientation

In this case the ranges on x and y are $$0 \leq y \leq 3$$ and $$0 \leq x \leq (3-y)/3$$. This gives another three possible correct answers.

 $$\displaystyle{\int_{0}^{2}{\int_{0}^{3}{\int_{0}^{(3-y)/3}{1~dx}~dy}~dz}}$$ $$\displaystyle{\int_{0}^{3}{\int_{0}^{2}{\int_{0}^{(3-y)/3}{1~dx}~dz}~dy}}$$ $$\displaystyle{\int_{0}^{3}{\int_{0}^{(3-y)/3}{\int_{0}^{2}{1~dz}~dx}~dy}}$$

(15 points) Set up (but do not evaluate) an integral in cylindrical coordinates to calculate the volume of the region bounded by the plane $$z=\sqrt{29}$$ and the hyperboloid $$z=\sqrt{4+x^2+y^2}$$.

Problem Statement

(15 points) Set up (but do not evaluate) an integral in cylindrical coordinates to calculate the volume of the region bounded by the plane $$z=\sqrt{29}$$ and the hyperboloid $$z=\sqrt{4+x^2+y^2}$$.

Solution

For starters, since we are asked to use cylindrical coordinates, we use the identity $$r^2=x^2+y^2$$ to get $$z=\sqrt{4+r^2}$$. Plotting the volume where it intersects in the yz-plane (x=0), we can find the z range as $$\sqrt{4+r^2} \leq z \leq \sqrt{29}$$. In the xy-plane, we have a circle of radius 5, so that $$0 \leq r \leq 5$$. This gives us the following three possible correct answers.

 $$\displaystyle{\int_{0}^{2\pi}{\int_{0}^{5}{\int_{\sqrt{4+r^2}}^{\sqrt{29}}{r~dz}~dr}~d\theta}}$$ $$\displaystyle{\int_{0}^{5}{\int_{0}^{2\pi}{\int_{\sqrt{4+r^2}}^{\sqrt{29}}{r~dz}~d\theta}~dr}}$$ $$\displaystyle{\int_{0}^{5}{\int_{\sqrt{4+r^2}}^{\sqrt{29}}{\int_{0}^{2\pi}{r~d\theta}~dz}~dr}}$$

(20 points) Set up (but do not evaluate) an integral in spherical coordinates to calculate the volume of the region inside the cone $$z=(x^2+y^2)^{1/2}$$ that lies between the planes $$z=1$$ and $$z=2$$.

Problem Statement

(20 points) Set up (but do not evaluate) an integral in spherical coordinates to calculate the volume of the region inside the cone $$z=(x^2+y^2)^{1/2}$$ that lies between the planes $$z=1$$ and $$z=2$$.

Solution

The plot above is the yz-plane trace (the intersection of the solid and the yz-plane). The angle lines are $$y=x$$ and $$y=-x$$. In spherical coordinates, these lines correlate with the angle $$\phi=\pi/4$$. The equations for $$z=1$$ and $$z=2$$ in spherical coordinates are $$\rho=\sec \phi$$ and $$\rho = 2\sec \phi$$ giving us the range on $$\phi$$ as $$\sec \phi \leq \rho \leq 2\sec \phi$$.

The range on $$\theta$$ is $$0 \leq \theta \leq 2\pi$$ and on $$\phi$$ we have $$0 \leq \phi \leq \pi/4$$. We can now set up the integral. There are three possible correct answers.

 $$\displaystyle{\int_{0}^{2\pi}{\int_{0}^{\pi/4}{\int_{\sec \phi}^{2\sec \phi}{\rho^2\sin\phi~d\rho}~d\phi}~d\theta}}$$ $$\displaystyle{\int_{0}^{\pi/4}{\int_{0}^{2\pi}{\int_{\sec \phi}^{2\sec \phi}{\rho^2\sin\phi~d\rho}~d\theta}~d\phi}}$$ $$\displaystyle{\int_{0}^{\pi/4}{\int_{\sec \phi}^{2\sec \phi}{\int_{0}^{2\pi}{\rho^2\sin\phi~d\theta}~d\rho}~d\phi}}$$