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learning and study techniques

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.

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

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

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.

GOT IT. THANKS!

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

Basic Problems

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

1950 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

\( 32/9 \)

Problem Statement

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

Solution

1951 solution video

video by Dr Chris Tisdell

Final Answer

\( 32/9 \)

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

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

1952 solution video

video by MIP4U

Final Answer

\( 9 \pi/2 \)

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

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

1953 solution video

video by MIP4U

Final Answer

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

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

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

1954 solution video

video by Krista King Math

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

\( \sqrt{2}/4 \)

Problem Statement

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

Solution

1957 solution video

video by MIT OCW

Final Answer

\( \sqrt{2}/4 \)

close solution

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

Problem Statement

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

1958 solution video

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

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

1959 solution video

video by MIT OCW

Final Answer

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

close solution

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

Problem Statement

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

Final Answer

\( 243 \pi/10 \)

Problem Statement

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

Solution

1960 solution video

video by MIP4U

Final Answer

\( 243 \pi/10 \)

close solution

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

1961 solution video

video by Krista King Math

Final Answer

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

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

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

1962 solution video

video by Krista King Math

Final Answer

\( 4 \pi/3 \)

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

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

1967 solution video

video by MIP4U

Final Answer

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

close solution

Intermediate Problems

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.

1955 solution video

video by PatrickJMT

1955 solution video

video by PatrickJMT

1955 solution video

video by PatrickJMT

Final Answer

\( 8 \pi \)

close solution

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

Problem Statement

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

Final Answer

\( 1/24 \)

Problem Statement

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

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).

1963 solution video

video by MIP4U

Final Answer

\( 1/24 \)

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

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

1964 solution video

video by MIP4U

Final Answer

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

close solution

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

1965 solution video

video by MIP4U

Final Answer

\( \pi/3 \)

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

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

1966 solution video

video by MIP4U

Final Answer

\( 169 \pi/10 \)

close solution
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