This is the third exam for third semester (multivariable) calculus. This exam contains 8 questions on partial integrals including double and triple integrals and their applications.
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IMPORTANT 
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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:
 Show all your work.
 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).
 Give exact, simplified answers.
(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 \(\displaystyle{ \int_{1}^{2}{ \int_{0}^{\ln(x)}{ x^3 e^y ~dy } ~dx } }\).
Final Answer 

\( 49/20 \)
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 ] 
\( \int_{1}^{2}{x^4  x^3~dx} \) 
\(\displaystyle{ \left[ \frac{x^5}{5}  \frac{x^4}{4} \right]_{1}^{2} }\) 
\( [2^5/5  2^4/4]  [1/5  1/4] = 49/20 \) 
Final Answer 

\( 49/20 \) 
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(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 \(y=2, y=2x\) and \(y=x2\).
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 \(y=2, y=2x\) and \(y=x2\).
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 \(2y \leq x \leq y+2\). So our integral is
\(\displaystyle{\int_{0}^{2}{\int_{2y}^{y+2}{ y^2 ~dx} ~dy}}\)
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(10 points) Sketch the region of integration and set up (but do not evaluate) this integral in polar coordinates \(\displaystyle{ \int_{5}^{5}{ \int_{0}^{\sqrt{25y^2}}{ 16x^2y^2 ~dx }~dy } }\)
Problem Statement 

(10 points) Sketch the region of integration and set up (but do not evaluate) this integral in polar coordinates \(\displaystyle{ \int_{5}^{5}{ \int_{0}^{\sqrt{25y^2}}{ 16x^2y^2 ~dx }~dy } }\)
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 \(0 \leq x \leq \sqrt{25y^2}\). 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.
\(\displaystyle{\int_{\pi/2}^{\pi/2}{\int_{0}^{5}{ (16r^2)~rdr}~d\theta}}\)
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. \(\displaystyle{\int_{0}^{5}{\int_{\pi/2}^{\pi/2}{(16r^2)~rd\theta}~dr}}\) is also a correct answer.
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(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 \(\sqrt{25x^2}\) to \(\sqrt{25x^2}\). The integrand does not change, so our answer is \(\displaystyle{ \int_{0}^{5}{ \int_{\sqrt{25x^2}}^{\sqrt{25x^2}}{ 16x^2y^2 ~ dy} ~dx } }\)
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(10 points) Evaluate \(\displaystyle{ \int_{0}^{\pi}{ \int_{0}^{y}{ \int_{0}^{\sin(x)}{ dz } ~dx } ~dy } }\)
Problem Statement 

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

\(\pi\)
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 ]
\(\displaystyle{\int_{0}^{\pi}{1\cos(y)~dy} = \left[ y\sin(y) \right]_{0}^{\pi} = \pi}\)
Final Answer 

\(\pi\) 
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(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 \(y=33x\) 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 \(y=33x\) 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 \(0 \leq y \leq 33x\). This gives these three possible correct answers.
\(\displaystyle{\int_{0}^{2}{\int_{0}^{1}{\int_{0}^{33x}{1~dy}~dx}~dz}}\) 
\(\displaystyle{\int_{0}^{1}{\int_{0}^{2}{\int_{0}^{33x}{1~dy}~dz}~dx}}\) 
\(\displaystyle{\int_{0}^{1}{\int_{0}^{33x}{\int_{0}^{2}{1~dz}~dy}~dx}}\) 
horizontal orientation 
In this case the ranges on x and y are \(0 \leq y \leq 3\) and \(0 \leq x \leq (3y)/3\). This gives another three possible correct answers.
\(\displaystyle{\int_{0}^{2}{\int_{0}^{3}{\int_{0}^{(3y)/3}{1~dx}~dy}~dz}}\) 
\(\displaystyle{\int_{0}^{3}{\int_{0}^{2}{\int_{0}^{(3y)/3}{1~dx}~dz}~dy}}\) 
\(\displaystyle{\int_{0}^{3}{\int_{0}^{(3y)/3}{\int_{0}^{2}{1~dz}~dx}~dy}}\) 
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(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 yzplane (x=0), we can find the z range as \(\sqrt{4+r^2} \leq z \leq \sqrt{29}\). In the xyplane, we have a circle of radius 5, so that \(0 \leq r \leq 5\). This gives us the following three possible correct answers.
\(\displaystyle{\int_{0}^{2\pi}{\int_{0}^{5}{\int_{\sqrt{4+r^2}}^{\sqrt{29}}{r~dz}~dr}~d\theta}}\) 
\(\displaystyle{\int_{0}^{5}{\int_{0}^{2\pi}{\int_{\sqrt{4+r^2}}^{\sqrt{29}}{r~dz}~d\theta}~dr}}\) 
\(\displaystyle{\int_{0}^{5}{\int_{\sqrt{4+r^2}}^{\sqrt{29}}{\int_{0}^{2\pi}{r~d\theta}~dz}~dr}}\) 
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(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 yzplane trace (the intersection of the solid and the yzplane). 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.
\(\displaystyle{\int_{0}^{2\pi}{\int_{0}^{\pi/4}{\int_{\sec \phi}^{2\sec \phi}{\rho^2\sin\phi~d\rho}~d\phi}~d\theta}}\) 
\(\displaystyle{\int_{0}^{\pi/4}{\int_{0}^{2\pi}{\int_{\sec \phi}^{2\sec \phi}{\rho^2\sin\phi~d\rho}~d\theta}~d\phi}}\) 
\(\displaystyle{\int_{0}^{\pi/4}{\int_{\sec \phi}^{2\sec \phi}{\int_{0}^{2\pi}{\rho^2\sin\phi~d\theta}~d\rho}~d\phi}}\) 
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You CAN Ace Calculus
other exams from calculus 3 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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