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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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Help Keep 17Calculus Free |
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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 xy-plane.
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 xy-plane 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
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on. |
Instructions - - Unless otherwise instructed, evaluate the following integrals using polar coordinates, giving your answers in exact terms.
Basic Problems |
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\(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1-y^2}}{ 3y~dx~dy} } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1-y^2}}{ 3y~dx~dy} } }\)
Final Answer |
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\( 1 - 1/\sqrt{2} \) |
Problem Statement |
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\(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1-y^2}}{ 3y~dx~dy} } }\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( 1 - 1/\sqrt{2} \) |
close solution |
\(\displaystyle{ \int_{0}^{2}{ \int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{2}{ \int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\)
Final Answer |
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\( 32/9 \) |
Problem Statement |
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\(\displaystyle{ \int_{0}^{2}{ \int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}{ \sqrt{x^2+y^2 }~dx~dy}} }\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( 32/9 \) |
close solution |
Determine the volume of the solid below the surface \( f(x,y) = 4 - x^2 - y^2 \) above the xy-plane over the region bounded by \( x^2 + y^2 = 1 \) and \( x^2 + y^2 = 4 \).
Problem Statement |
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Determine the volume of the solid below the surface \( f(x,y) = 4 - x^2 - y^2 \) above the xy-plane over the region bounded by \( x^2 + y^2 = 1 \) and \( x^2 + y^2 = 4 \).
Final Answer |
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\( 9 \pi/2 \) |
Problem Statement |
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Determine the volume of the solid below the surface \( f(x,y) = 4 - x^2 - y^2 \) above the xy-plane over the region bounded by \( x^2 + y^2 = 1 \) and \( x^2 + y^2 = 4 \).
Solution |
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video by MIP4U
Final Answer |
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\( 9 \pi/2 \) |
close solution |
Determine the volume of \( z = \sqrt{9-x^2-y^2} \) over the region \( x^2 + y^2 \leq 4 \) in the first octant.
Problem Statement |
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Determine the volume of \( z = \sqrt{9-x^2-y^2} \) over the region \( x^2 + y^2 \leq 4 \) in the first octant.
Final Answer |
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\( (27-5\sqrt{5}) \pi/6 \) |
Problem Statement |
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Determine the volume of \( z = \sqrt{9-x^2-y^2} \) over the region \( x^2 + y^2 \leq 4 \) in the first octant.
Solution |
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video by MIP4U
Final Answer |
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\( (27-5\sqrt{5}) \pi/6 \) |
close solution |
Evaluate \( \iint_{A}{ e^{-x^2-y^2} ~dA } \) where A is bounded by \( x = \sqrt{4-y^2} \) and the y-axis.
Problem Statement |
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Evaluate \( \iint_{A}{ e^{-x^2-y^2} ~dA } \) where A is bounded by \( x = \sqrt{4-y^2} \) and the y-axis.
Final Answer |
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\( (\pi/2)(1-1/e^4) \) |
Problem Statement |
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Evaluate \( \iint_{A}{ e^{-x^2-y^2} ~dA } \) where A is bounded by \( x = \sqrt{4-y^2} \) and the y-axis.
Solution |
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video by Krista King Math
Final Answer |
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\( (\pi/2)(1-1/e^4) \) |
close solution |
\(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\)
Problem Statement |
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\(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\)
Final Answer |
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\( \sqrt{2}/4 \) |
Problem Statement |
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\(\displaystyle{ \int_{x=1}^{2}{ \int_{y=0}^{x}{\frac{1}{ (x^2+y^2)^{3/2}}~dy~dx } } }\)
Solution |
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video by MIT OCW
Final Answer |
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\( \sqrt{2}/4 \) |
close solution |
Convert the integral \(\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }\) to polar coordinates.
Problem Statement |
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Convert the integral \(\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }\) to polar coordinates.
Final Answer |
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\(\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }\) |
Problem Statement |
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Convert the integral \(\displaystyle{ \int_{x=0}^{1}{ \int_{y=x^2}^{x}{ f~dy~dx } } }\) to polar coordinates.
Solution |
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video by MIT OCW
Final Answer |
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\(\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }\) |
close solution |
Convert \(\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2y-y^2}}{ f~dx~dy } } }\) to polar coordinates.
Problem Statement |
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Convert \(\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2y-y^2}}{ f~dx~dy } } }\) to polar coordinates.
Final Answer |
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\(\displaystyle{ \int_{\theta=0}^{\pi/2}{ \int_{r=0}^{2\sin\theta}{ f~rdr~d\theta } } }\) |
Problem Statement |
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Convert \(\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2y-y^2}}{ f~dx~dy } } }\) to polar coordinates.
Solution |
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video by MIT OCW
Final Answer |
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\(\displaystyle{ \int_{\theta=0}^{\pi/2}{ \int_{r=0}^{2\sin\theta}{ f~rdr~d\theta } } }\) |
close solution |
\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\)
Final Answer |
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\( 243 \pi/10 \) |
Problem Statement |
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\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\)
Solution |
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video by MIP4U
Final Answer |
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\( 243 \pi/10 \) |
close solution |
\(\displaystyle{ \int_{-3}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sin(x^2+y^2)~dy~dx } } }\)
Problem Statement |
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\(\displaystyle{ \int_{-3}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sin(x^2+y^2)~dy~dx } } }\)
Final Answer |
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\( (1-\cos 9) \pi/2 \) |
Problem Statement |
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\(\displaystyle{ \int_{-3}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sin(x^2+y^2)~dy~dx } } }\)
Solution |
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video by Krista King Math
Final Answer |
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\( (1-\cos 9) \pi/2 \) |
close solution |
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 |
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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 \( z = 2 \).
Final Answer |
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\( 4 \pi/3 \) |
Problem Statement |
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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 \( z = 2 \).
Solution |
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video by Krista King Math
Final Answer |
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\( 4 \pi/3 \) |
close solution |
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 |
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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.
Final Answer |
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\( (4\pi/3)(24)^{3/2} \) cm^{3} |
Problem Statement |
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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.
Solution |
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video by MIP4U
Final Answer |
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\( (4\pi/3)(24)^{3/2} \) cm^{3} |
close solution |
Intermediate Problems |
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Calculate the area under the plane \( 6x + 4y + z = 12 \) above the disk with boundary circle \( x^2 + y^2 = 2y \).
Problem Statement |
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Calculate the area under the plane \( 6x + 4y + z = 12 \) above the disk with boundary circle \( x^2 + y^2 = 2y \).
Final Answer |
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\( 8 \pi \) |
Problem Statement |
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Calculate the area under the plane \( 6x + 4y + z = 12 \) above the disk with boundary circle \( x^2 + y^2 = 2y \).
Solution |
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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 |
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\( 8 \pi \) |
close solution |
\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{x-x^2}}{ xy ~dy ~dx } } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{x-x^2}}{ xy ~dy ~dx } } }\)
Final Answer |
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\( 1/24 \) |
Problem Statement |
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\(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{x-x^2}}{ xy ~dy ~dx } } }\)
Solution |
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About 2 minutes and 30 seconds into this video he builds what he calls a t-table 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 semi-circle of radius \(1/2\) and center at \((1/2,0)\). Here is how we would do it.
We know that \(0\leq y\leq\sqrt{x-x^2}\). This means that the boundary is \(y=\sqrt{x-x^2}\) and \(y\geq 0\). So we complete the square on the boundary equation, as follows.
\( y = \sqrt{x-x^2} \) |
\( y^2 = x-x^2 \) |
\( 0 = x^2-x+y^2 \) |
\( 0 = x^2-x+1/4 -1/4 +y^2 \) |
\( 1/4 = (x-1/2)^2 + y^2 \) |
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 |
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\( 1/24 \) |
close solution |
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 |
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Find the surface area of the part of the plane \( z = 3 + 2x + 4y \) that lies inside the cylinder \( x^2 + y^2 = 4 \).
Final Answer |
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\( 4 \pi \sqrt{21} \) |
Problem Statement |
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Find the surface area of the part of the plane \( z = 3 + 2x + 4y \) that lies inside the cylinder \( x^2 + y^2 = 4 \).
Solution |
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video by MIP4U
Final Answer |
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\( 4 \pi \sqrt{21} \) |
close solution |
Determine the area of one petal of \( r = 2 \sin(3\theta) \).
Problem Statement |
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Determine the area of one petal of \( r = 2 \sin(3\theta) \).
Final Answer |
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\( \pi/3 \) |
Problem Statement |
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Determine the area of one petal of \( r = 2 \sin(3\theta) \).
Solution |
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video by MIP4U
Final Answer |
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\( \pi/3 \) |
close solution |
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 |
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Find the volume of the solid bounded by the paraboloids \( z = -6 + 3x^2 + 3y^2 \) and \( z = 7 - 2x^2 - 2y^2 \).
Final Answer |
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\( 169 \pi/10 \) |
Problem Statement |
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Find the volume of the solid bounded by the paraboloids \( z = -6 + 3x^2 + 3y^2 \) and \( z = 7 - 2x^2 - 2y^2 \).
Solution |
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video by MIP4U
Final Answer |
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\( 169 \pi/10 \) |
close solution |