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You CAN Ace Calculus

17calculus > partial integrals > double integrals - polar

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double integrals in polar coordinates youtube playlist

Double Integrals in Polar Coordinates

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 (51min-29secs)
 Evans Lawrence - Multivariable Calculus: Lecture 19 - Double Integration in Polar Coordinates (35min-2secs)

### Search 17Calculus

Practice Problems

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

 Level A - Basic

Practice A01

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

solution

Practice A02

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

solution

Practice A03

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

Practice A04

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

Practice A05

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

solution

Practice A06

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

solution

Practice A07

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

solution

Practice A08

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

solution

Practice A09

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

solution

Practice A10

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

solution

Practice A11

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

Practice A12

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

 Level B - Intermediate

Practice B01

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

solution

Practice B02

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

solution

Practice B03

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

Practice B04

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

Find the volume of the solid bounded by the paraboloids $$z=-6+3x^2+3y^2$$ and $$z=7-2x^2-2y^2$$.