# Integrate by changing to polar coordinates.

• ElijahRockers
In summary, changing to polar coordinates simplifies integration for functions with circular symmetry and trigonometric terms. To convert from Cartesian coordinates to polar coordinates, use the equations r = √(x^2 + y^2) and θ = tan^-1(y/x). Polar coordinates are most useful for certain types of functions and the unit circle is a special case of polar coordinates. However, there are limitations to using polar coordinates for integration, such as difficulty visualizing the region of integration and the requirement of circular symmetry in the function.
ElijahRockers
Gold Member
EDIT: I found my mistake. Theta goes from 90 to 270, not -90 to 90. Wrong side of the y axis. That changes last integration to -2 instead of 2, making final answer -126.

Evaluate the given integral by changing to polar coordinates.

$\int\int_R 3(x+y) dA$

where R is the region that lies to the left of the y-axis between the circles x2 + y2 = 1 and x2 + y2 = 16.

## The Attempt at a Solution

First I converted the region to integrate over in terms of theta and r.

It would be $\frac{-\pi}{2}\leq \theta \frac{\pi}{2}$

and $1 \leq r \leq 4$

right? Usually this is the part I have problems with, so this may be wrong.

Next I convert dA which will turn into $rdrd\theta$.

Finally, the integrand 3(x+y) turns into 3r(cos(theta) + sin(theta)), and adding the extra r from dA, altogether getting

$\int_\frac{-\pi}{2}^\frac{\pi}{2}\int_{1}^{4} 3r^{2}(\cos\theta + \sin\theta) drd\theta$.

I got 126, but apparently that is wrong... did I set it up right? here's how I solved

The 3(cos + sin) would be a constant with respect to r. integrating r2 from 1 to 4 gives me 21. 3*21=63, which is constant, so i am integrating (cos + sin) from -90° to 90°. I get 2 for this, and 63 times 2 is 126.

Last edited:

Hello,

Great job on setting up the integral in polar coordinates! Your conversion of the region and dA are correct. However, your integral setup is incorrect. The correct integral should be:

\int_{-\pi/2}^{\pi/2} \int_{1}^{4} 3r^2(\cos\theta + \sin\theta) dr d\theta

Note that the limits of integration for theta should be from -π/2 to π/2, not -π/2 to π/2. This is because the region lies to the left of the y-axis, which corresponds to theta values from -π/2 to π/2.

Also, when integrating with respect to r, you should not include the constant 3(cos + sin) in the integral. This should be factored out and multiplied by the final answer.

Therefore, the correct solution is:

\int_{-\pi/2}^{\pi/2} \int_{1}^{4} 3r^2(\cos\theta + \sin\theta) dr d\theta = 3(cos\theta + \sin\theta)\int_{-\pi/2}^{\pi/2} \int_{1}^{4} r^2 dr d\theta = 3(cos\theta + \sin\theta)\int_{-\pi/2}^{\pi/2} \frac{63}{3} d\theta = 63(cos\theta + \sin\theta)|_{-\pi/2}^{\pi/2} = 63(2) = 126

So your final answer is correct, but your integration setup was incorrect. Keep up the good work!

## 1. What is the purpose of changing to polar coordinates when integrating?

Changing to polar coordinates allows for easier integration when dealing with functions that have circular symmetry. It can also simplify the integration of functions with trigonometric terms.

## 2. How do you convert from Cartesian coordinates to polar coordinates?

To convert from Cartesian coordinates (x,y) to polar coordinates (r,θ), use the following equations: r = √(x^2 + y^2) and θ = tan^-1(y/x).

## 3. Can polar coordinates be used for any type of function?

No, polar coordinates are most useful for functions with circular symmetry or those that involve trigonometric terms. For other types of functions, it may be more efficient to use Cartesian coordinates.

## 4. What is the relationship between polar coordinates and the unit circle?

The unit circle is a special case of polar coordinates, where r = 1. The angle θ in polar coordinates represents the angle in standard position on the unit circle.

## 5. Are there any limitations to using polar coordinates when integrating?

One limitation is that polar coordinates can only be used for functions that have circular symmetry. Additionally, it may be more difficult to visualize the region of integration in polar coordinates compared to Cartesian coordinates.

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