This page covers double integrals in polar coordinates. Double integrals in rectangular coordinates are covered on a separate page.
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 

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

Problem Statement 

\(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1y^2}}{ 3y~dx~dy} } }\)
Final Answer 

Problem Statement 

\(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1y^2}}{ 3y~dx~dy} } }\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( 1  1/\sqrt{2} \) 
close solution

Problem Statement 

\(\displaystyle{ \int_{0}^{2}{ \int_{\sqrt{2yy^2}}^{\sqrt{2yy^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\)
Final Answer 

Problem Statement 

\(\displaystyle{ \int_{0}^{2}{ \int_{\sqrt{2yy^2}}^{\sqrt{2yy^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\)
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( 32/9 \) 
close solution

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 

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 

\( 9 \pi/2 \) 
close solution

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 

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 

\( (275\sqrt{5}) \pi/6 \) 
close solution

Problem Statement 

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

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 

\( (\pi/2)(11/e^4) \) 
close solution

Problem Statement 

\(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\)
Final Answer 

Problem Statement 

\(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\)
Solution 

video by MIT OCW 

Final Answer 

\( \sqrt{2}/4 \) 
close solution

Problem Statement 

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

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 

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

Problem Statement 

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

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 

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

Problem Statement 

\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\)
Final Answer 

Problem Statement 

\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\)
Solution 

video by MIP4U 

Final Answer 

\( 243 \pi/10 \) 
close solution

Problem Statement 

\(\displaystyle{ \int_{3}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sin(x^2+y^2)~dy~dx } } }\)
Final Answer 

Problem Statement 

\(\displaystyle{ \int_{3}^{3}{ \int_{0}^{\sqrt{9x^2}}{ \sin(x^2+y^2)~dy~dx } } }\)
Solution 

video by Krista King Math 

Final Answer 

\( (1\cos 9) \pi/2 \) 
close solution

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 

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 

\( 4 \pi/3 \) 
close solution

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 

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} 
close solution

Intermediate Problems 

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 

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 

\( 8 \pi \) 
close solution

Problem Statement 

\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{xx^2}}{ xy ~dy ~dx } } }\)
Final Answer 

Problem Statement 

\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{xx^2}}{ xy ~dy ~dx } } }\)
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 

\( 1/24 \) 
close solution

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 

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 

\( 4 \pi \sqrt{21} \) 
close solution

Problem Statement 

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

Problem Statement 

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

video by MIP4U 

Final Answer 

\( \pi/3 \) 
close solution

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 

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 

\( 169 \pi/10 \) 
close solution

Here is a playlist of the videos on this page.
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