This page covers double integrals in polar coordinates.
As you learned on the polar coordinates page, you use the equations \(x=r\cos\theta\) and \(y=r\sin\theta\) to convert equations from rectangular to polar coordinates. The same idea applies to a function and the description of an area in the xyplane.
For example, if you have an integral in rectangular coordinates that looks like \(\iint_A{f(x,y)~dA}\), you need to do three things to convert this to polar coordinates.
1. First, substitute for x and y in \(f(x,y)\) using the above equations to get \(f(r\cos\theta,r\sin\theta)\). This new form of the function can be written as \(f(r,\theta)\).
2. Second, describe the area in polar coordinates or, if the area is already given in rectangular coordinates, convert the area in the xyplane from rectangular to polar coordinates.
3. Finally, set up the integral with the function \(f(r,\theta)\) in polar coordinates, being careful to integrate in the correct order.
Okay, so that is the big picture but how do you implement this when working problems? Time for some videos. Both of these videos are rather long but they will give you a good handle on double integrals in polar coordinates. You do not need to watch both of them since they cover the same ideas. But if you do, you will have a better understanding of the techniques.
video by MIT OCW 

video by Evans Lawrence 

Practice
Unless otherwise instructed, evaluate the following integrals using polar coordinates, giving your answers in exact terms.
Basic 

Calculate the surface area of the part of \( z = x^2 + y^2 \) below the plane \( z=9 \).
Problem Statement 

Calculate the surface area of the part of \( z = x^2 + y^2 \) below the plane \( z=9 \).
Final Answer 

\(\displaystyle{ \frac{\pi}{6}\left[ 37^{3/2} 1 \right] }\)
Problem Statement 

Calculate the surface area of the part of \( z = x^2 + y^2 \) below the plane \( z=9 \).
Solution 

The instructor does not show how to evaluate the final answer. They just give the answer. Even though the evaluation is pretty simple, here are the details.
\(\displaystyle{ \int_{0}^{2\pi}{ \int_{0}^{3}{ \sqrt{4r^2+1} \cdot r~dr~d\theta } } }\) 
Let's work on the inside integral first. Let \( u = 4r^2 + 1 \to du = 8r~dr \) 
For the limits of integration, \( r=0 \to u=1 \), \( r=3 \to u = 4(9)+1 =37 \) 
\(\displaystyle{ \int_{0}^{3}{ \sqrt{4r^2+1} \cdot r~dr } }\) 
\(\displaystyle{ \int_{1}^{37}{ u^{1/2} \frac{du}{8} } }\) 
\(\displaystyle{ \left[ \frac{u^{3/2}}{12} \right]_{1}^{37} }\) 
So the inside integral evaluates to \(\displaystyle{ \frac{1}{12}\left[ 37^{3/2}  1 \right] }\) 
So now we have \(\displaystyle{ \int_{0}^{2\pi}{ \frac{1}{12}\left[ 37^{3/2}  1 \right] ~ d\theta } }\) 
Since there are no \(\theta\)'s in the integrand, we end up with \(\displaystyle{ \frac{2\pi}{12}\left[ 37^{3/2}  1 \right] }\) 
which simplifies to \(\displaystyle{ \frac{\pi}{6}\left[ 37^{3/2} 1 \right] }\) 
video by Alexandra Budden 

Final Answer 

\(\displaystyle{ \frac{\pi}{6}\left[ 37^{3/2} 1 \right] }\) 
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1y^2}}{ 3y~dx~dy} } }\) using polar coordinates, giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1y^2}}{ 3y~dx~dy} } }\) using polar coordinates, giving your answer in exact terms.
Final Answer 

\( 1  1/\sqrt{2} \)
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1y^2}}{ 3y~dx~dy} } }\) using polar coordinates, giving your answer in exact terms.
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( 1  1/\sqrt{2} \) 
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{2}{ \int_{\sqrt{2yy^2}}^{\sqrt{2yy^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\) using polar coordinates, giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{2}{ \int_{\sqrt{2yy^2}}^{\sqrt{2yy^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\) using polar coordinates, giving your answer in exact terms.
Final Answer 

\( 32/9 \)
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{2}{ \int_{\sqrt{2yy^2}}^{\sqrt{2yy^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\) using polar coordinates, giving your answer in exact terms.
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( 32/9 \) 
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Determine the volume of the solid below the surface \( f(x,y) = 4  x^2  y^2 \) above the xyplane over the region bounded by \( x^2 + y^2 = 1 \) and \( x^2 + y^2 = 4 \).
Problem Statement 

Determine the volume of the solid below the surface \( f(x,y) = 4  x^2  y^2 \) above the xyplane over the region bounded by \( x^2 + y^2 = 1 \) and \( x^2 + y^2 = 4 \).
Final Answer 

\( 9 \pi/2 \)
Problem Statement 

Determine the volume of the solid below the surface \( f(x,y) = 4  x^2  y^2 \) above the xyplane over the region bounded by \( x^2 + y^2 = 1 \) and \( x^2 + y^2 = 4 \).
Solution 

video by MIP4U 

Final Answer 

\( 9 \pi/2 \) 
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Determine the volume of \( z = \sqrt{9x^2y^2} \) over the region \( x^2 + y^2 \leq 4 \) in the first octant.
Problem Statement 

Determine the volume of \( z = \sqrt{9x^2y^2} \) over the region \( x^2 + y^2 \leq 4 \) in the first octant.
Final Answer 

\( (275\sqrt{5}) \pi/6 \)
Problem Statement 

Determine the volume of \( z = \sqrt{9x^2y^2} \) over the region \( x^2 + y^2 \leq 4 \) in the first octant.
Solution 

video by MIP4U 

Final Answer 

\( (275\sqrt{5}) \pi/6 \) 
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Evaluate \( \iint_{A}{ e^{x^2y^2} ~dA } \) where A is bounded by \( x = \sqrt{4y^2} \) and the yaxis.
Problem Statement 

Evaluate \( \iint_{A}{ e^{x^2y^2} ~dA } \) where A is bounded by \( x = \sqrt{4y^2} \) and the yaxis.
Final Answer 

\( (\pi/2)(11/e^4) \)
Problem Statement 

Evaluate \( \iint_{A}{ e^{x^2y^2} ~dA } \) where A is bounded by \( x = \sqrt{4y^2} \) and the yaxis.
Solution 

video by Krista King Math 

Final Answer 

\( (\pi/2)(11/e^4) \) 
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Final Answer 

\( \sqrt{2}/4 \)
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Solution 

video by MIT OCW 

Final Answer 

\( \sqrt{2}/4 \) 
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Convert the integral \(\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }\) to polar coordinates.
Problem Statement 

Convert the integral \(\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }\) to polar coordinates.
Final Answer 

\(\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }\)
Problem Statement 

Convert the integral \(\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }\) to polar coordinates.
Solution 

video by MIT OCW 

Final Answer 

\(\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }\) 
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Convert \(\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2yy^2}}{ f~dx~dy } } }\) to polar coordinates.
Problem Statement 

Convert \(\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2yy^2}}{ f~dx~dy } } }\) to polar coordinates.
Final Answer 

\(\displaystyle{ \int_{\theta=0}^{\pi/2}{ \int_{r=0}^{2\sin\theta}{ f~rdr~d\theta } } }\)
Problem Statement 

Convert \(\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2yy^2}}{ f~dx~dy } } }\) to polar coordinates.
Solution 

video by MIT OCW 

Final Answer 

\(\displaystyle{ \int_{\theta=0}^{\pi/2}{ \int_{r=0}^{2\sin\theta}{ f~rdr~d\theta } } }\) 
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Final Answer 

\( 243 \pi/10 \)
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Solution 

video by MIP4U 

Final Answer 

\( 243 \pi/10 \) 
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{3}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sin(x^2+y^2)~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{3}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sin(x^2+y^2)~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Final Answer 

\( (1\cos 9) \pi/2 \)
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{3}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sin(x^2+y^2)~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.
Solution 

video by Krista King Math 

Final Answer 

\( (1\cos 9) \pi/2 \) 
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Use a double polar integral to find the volume of the solid enclosed by \( x^2  y^2 + z^2 = 1 \) and \( z = 2 \).
Problem Statement 

Use a double polar integral to find the volume of the solid enclosed by \( x^2  y^2 + z^2 = 1 \) and \( z = 2 \).
Final Answer 

\( 4 \pi/3 \)
Problem Statement 

Use a double polar integral to find the volume of the solid enclosed by \( x^2  y^2 + z^2 = 1 \) and \( z = 2 \).
Solution 

video by Krista King Math 

Final Answer 

\( 4 \pi/3 \) 
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A cylindrical drill with a radius of 5 cm is used to bore a hole through the center of a sphere of radius 7 cm. Find the volume of the ring shaped solid that remains.
Problem Statement 

A cylindrical drill with a radius of 5 cm is used to bore a hole through the center of a sphere of radius 7 cm. Find the volume of the ring shaped solid that remains.
Final Answer 

\( (4\pi/3)(24)^{3/2} \) cm^{3}
Problem Statement 

A cylindrical drill with a radius of 5 cm is used to bore a hole through the center of a sphere of radius 7 cm. Find the volume of the ring shaped solid that remains.
Solution 

video by MIP4U 

Final Answer 

\( (4\pi/3)(24)^{3/2} \) cm^{3} 
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Intermediate 

Calculate the area under the plane \( 6x + 4y + z = 12 \) above the disk with boundary circle \( x^2 + y^2 = 2y \).
Problem Statement 

Calculate the area under the plane \( 6x + 4y + z = 12 \) above the disk with boundary circle \( x^2 + y^2 = 2y \).
Final Answer 

\( 8 \pi \)
Problem Statement 

Calculate the area under the plane \( 6x + 4y + z = 12 \) above the disk with boundary circle \( x^2 + y^2 = 2y \).
Solution 

This problem is solved in three videos. In the first video he sets up the integral. In the second and third videos he evaluates the integral.
video by PatrickJMT 

video by PatrickJMT 

video by PatrickJMT 

Final Answer 

\( 8 \pi \) 
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{xx^2}}{ xy ~dy ~dx } } }\) using polar coordinates, giving your answer in exact terms.
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{xx^2}}{ xy ~dy ~dx } } }\) using polar coordinates, giving your answer in exact terms.
Final Answer 

\( 1/24 \)
Problem Statement 

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{xx^2}}{ xy ~dy ~dx } } }\) using polar coordinates, giving your answer in exact terms.
Solution 

About 2 minutes and 30 seconds into this video he builds what he calls a ttable to try to figure out what boundary of the region looks like. We would probably, instead, try to complete the square to get an equation for the boundary. His way of doing it does not really show that this is a semicircle of radius \(1/2\) and center at \((1/2,0)\). Here is how we would do it.
We know that \(0\leq y\leq\sqrt{xx^2}\). This means that the boundary is \(y=\sqrt{xx^2}\) and \(y\geq 0\). So we complete the square on the boundary equation, as follows.
\( y = \sqrt{xx^2} \) 
\( y^2 = xx^2 \) 
\( 0 = x^2x+y^2 \) 
\( 0 = x^2x+1/4 1/4 +y^2 \) 
\( 1/4 = (x1/2)^2 + y^2 \) 
In going from line 1 to line 2, we squared both sides of the equation. This is valid for x and y since \(y\geq 0\).
We would not recommend doing it the way he did in the video, since it is so inexact. However, a graphing calculator or app would help (but don't rely on them too much).
video by MIP4U 

Final Answer 

\( 1/24 \) 
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Find the surface area of the part of the plane \( z = 3 + 2x + 4y \) that lies inside the cylinder \( x^2 + y^2 = 4 \).
Problem Statement 

Find the surface area of the part of the plane \( z = 3 + 2x + 4y \) that lies inside the cylinder \( x^2 + y^2 = 4 \).
Final Answer 

\( 4 \pi \sqrt{21} \)
Problem Statement 

Find the surface area of the part of the plane \( z = 3 + 2x + 4y \) that lies inside the cylinder \( x^2 + y^2 = 4 \).
Solution 

video by MIP4U 

Final Answer 

\( 4 \pi \sqrt{21} \) 
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Determine the area of one petal of \( r = 2 \sin(3\theta) \).
Problem Statement 

Determine the area of one petal of \( r = 2 \sin(3\theta) \).
Final Answer 

\( \pi/3 \)
Problem Statement 

Determine the area of one petal of \( r = 2 \sin(3\theta) \).
Solution 

video by MIP4U 

Final Answer 

\( \pi/3 \) 
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Find the volume of the solid bounded by the paraboloids \( z = 6 + 3x^2 + 3y^2 \) and \( z = 7  2x^2  2y^2 \).
Problem Statement 

Find the volume of the solid bounded by the paraboloids \( z = 6 + 3x^2 + 3y^2 \) and \( z = 7  2x^2  2y^2 \).
Final Answer 

\( 169 \pi/10 \)
Problem Statement 

Find the volume of the solid bounded by the paraboloids \( z = 6 + 3x^2 + 3y^2 \) and \( z = 7  2x^2  2y^2 \).
Solution 

video by MIP4U 

Final Answer 

\( 169 \pi/10 \) 
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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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Practice Instructions
Unless otherwise instructed, evaluate the following integrals using polar coordinates, giving your answers in exact terms.