## 17Calculus - 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.

video by MIT OCW

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

video by Evans Lawrence

### Practice

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

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

$$1 - 1/\sqrt{2}$$

Problem Statement

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

Solution

### 1950 video

video by Dr Chris Tisdell

$$1 - 1/\sqrt{2}$$

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

Problem Statement

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

$$32/9$$

Problem Statement

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

Solution

### 1951 video

video by Dr Chris Tisdell

$$32/9$$

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 $$x^2 + y^2 = 4$$.

$$9 \pi/2$$

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 video

video by MIP4U

$$9 \pi/2$$

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.

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

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 video

video by MIP4U

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

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.

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

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 video

video by Krista King Math

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

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

Problem Statement

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

$$\sqrt{2}/4$$

Problem Statement

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

Solution

### 1957 video

video by MIT OCW

$$\sqrt{2}/4$$

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.

$$\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }$$

Problem Statement

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

Solution

### 1958 video

video by MIT OCW

$$\displaystyle{ \int_{\theta=0}^{\pi/4}{ \int_{r=0}^{\tan\theta\sec\theta}{ f~rdr~d\theta } } }$$

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.

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

Problem Statement

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

Solution

### 1959 video

video by MIT OCW

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

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

Problem Statement

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

$$243 \pi/10$$

Problem Statement

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

Solution

### 1960 video

video by MIP4U

$$243 \pi/10$$

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

Problem Statement

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

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

Problem Statement

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

Solution

### 1961 video

video by Krista King Math

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

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 $$z = 2$$.

$$4 \pi/3$$

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 video

video by Krista King Math

$$4 \pi/3$$

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.

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

video by MIP4U

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

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 $$x^2 + y^2 = 2y$$.

$$8 \pi$$

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 video

video by PatrickJMT

### 1955 video

video by PatrickJMT

### 1955 video

video by PatrickJMT

$$8 \pi$$

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

Problem Statement

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

$$1/24$$

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 video

video by MIP4U

$$1/24$$

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 $$x^2 + y^2 = 4$$.

$$4 \pi \sqrt{21}$$

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 video

video by MIP4U

$$4 \pi \sqrt{21}$$

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

Problem Statement

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

$$\pi/3$$

Problem Statement

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

Solution

### 1965 video

video by MIP4U

$$\pi/3$$

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 $$z = 7 - 2x^2 - 2y^2$$.

$$169 \pi/10$$

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 video

video by MIP4U

$$169 \pi/10$$

### double integrals in polar coordinates 17calculus youtube playlist

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