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17Calculus - Double Integrals in Polar Coordinates

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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 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.

MIT OCW - Lec 17 | MIT 18.02 Multivariable Calculus, Fall 2007 [51mins-29secs]

video by MIT OCW

Evans Lawrence - Multivariable Calculus: Lecture 19 - Double Integration in Polar Coordinates [35mins-2secs]

video by Evans Lawrence

Calculus, Better Explained: A Guide To Developing Lasting Intuition

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

Alexandra Budden - 3539 video solution

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 \(\displaystyle{ \int_{0}^{1/\sqrt{2}}{ \int_{y}^{\sqrt{1-y^2}}{ 3y~dx~dy} } }\) using polar coordinates, giving your answer in exact terms.

Problem Statement

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

Final Answer

\( 1 - 1/\sqrt{2} \)

Problem Statement

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

Solution

Dr Chris Tisdell - 1950 video solution

video by Dr Chris Tisdell

Final Answer

\( 1 - 1/\sqrt{2} \)

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

Problem Statement

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

Final Answer

\( 32/9 \)

Problem Statement

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

Solution

Dr Chris Tisdell - 1951 video 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 xy-plane 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 xy-plane 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 xy-plane over the region bounded by \( x^2 + y^2 = 1 \) and \( x^2 + y^2 = 4 \).

Solution

MIP4U - 1952 video solution

video by MIP4U

Final Answer

\( 9 \pi/2 \)

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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.

Problem Statement

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

\( (27-5\sqrt{5}) \pi/6 \)

Problem Statement

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

MIP4U - 1953 video solution

video by MIP4U

Final Answer

\( (27-5\sqrt{5}) \pi/6 \)

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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.

Problem Statement

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

\( (\pi/2)(1-1/e^4) \)

Problem Statement

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

Solution

Krista King Math - 1954 video solution

video by Krista King Math

Final Answer

\( (\pi/2)(1-1/e^4) \)

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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

MIT OCW - 1957 video 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

MIT OCW - 1958 video 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 \(\displaystyle{ \int_{y=0}^{2}{ \int_{x=0}^{\sqrt{2y-y^2}}{ f~dx~dy } } }\) to polar coordinates.

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

MIT OCW - 1959 video 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 \(\displaystyle{ \int_{0}^{3}{ \int_{0}^{\sqrt{9-x^2}}{ \sqrt{(x^2+y^2)^3 }~dy~dx } } }\) using polar coordinates, giving your answer in exact terms.

Problem Statement

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

Final Answer

\( 243 \pi/10 \)

Problem Statement

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

Solution

MIP4U - 1960 video solution

video by MIP4U

Final Answer

\( 243 \pi/10 \)

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

Problem Statement

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

Final Answer

\( (1-\cos 9) \pi/2 \)

Problem Statement

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

Solution

Krista King Math - 1961 video 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

Krista King Math - 1962 video 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

\( (4\pi/3)(24)^{3/2} \) cm3

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

MIP4U - 1967 video solution

video by MIP4U

Final Answer

\( (4\pi/3)(24)^{3/2} \) cm3

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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.

PatrickJMT - 1955 video solution

video by PatrickJMT

PatrickJMT - 1955 video solution

video by PatrickJMT

PatrickJMT - 1955 video solution

video by PatrickJMT

Final Answer

\( 8 \pi \)

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

Problem Statement

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

Final Answer

\( 1/24 \)

Problem Statement

Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{0}^{1}{ \int_{0}^{\sqrt{x-x^2}}{ xy ~dy ~dx } } }\) 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 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).

MIP4U - 1963 video solution

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

MIP4U - 1964 video 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

MIP4U - 1965 video 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

MIP4U - 1966 video solution

video by MIP4U

Final Answer

\( 169 \pi/10 \)

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Practice Instructions

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

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