# Homework Help: Convert this integral from cartesian coordinates to polar coordinates

1. Nov 14, 2014

### s3a

1. The problem statement, all variables and given/known data
The problem and its solution are attached as TheProblemAndTheSolution.jpg.

If you don't want to view the attached image, the cartesian-coordinate version that the problem wants me to convert to a polar-coordinate version is the following (let "int" = "integral").:
int int (1 - x^2 - y^2) dx from 0 to sqrt(2y - y^2) dy from 0 to 1

2. Relevant equations
Let t = theta (because it is simpler to type). Then x = rcos(t), y = rcos(t), r^2 = x^2 + y^2

3. The attempt at a solution
I completed the square and got this circle ( http://www.wolframalpha.com/input/?i=plot x^2 + (y - 1)^2 = 1 ).

Let "int" = "integral".

I thought I was supposed to do int int (1 - r^2) r dr from 0 to 1 (because the radius ranges in size from 0 to 1) dt from 0 to pi (because, I was initially thinking 0 to -pi, but then I thought about the circle being above the x axis and that the area would be positive).

Why does the solution attached get what it gets?

#### Attached Files:

• ###### TheProblemAndSolution.jpg
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Last edited: Nov 14, 2014
2. Nov 15, 2014

### Staff: Mentor

Both of your sets of integration limits are wrong.
In the Cartesian integral, x ranges between 0 and $\sqrt{2y - y^2}$. This means that you are not going to get the complete circle. This will have an effect on the range of values for $\theta$ in the polar integral.

Also, in the polar integral, r does not range from 0 to 1 as you say above. The values for r range between 0 and the partial circle. As you have the limits, the circle would be centered at (0, 0), which is incorrect.

3. Nov 22, 2014

### s3a

Sorry for the delayed response.

I'm still very confused as to how I am supposed to get (letting t = theta) the drawing with r = 2sin(t) and r = csc(t).

Could you please explain that part to me?

4. Nov 22, 2014

### Staff: Mentor

The Cartesian limits of integration define a region that is a quarter of a circle (the lower right quadrant). The center of the circle is at (0, 1) and its radius is 1. When you switch to polar form you will need two integrals, one in which r ranges from 0 to the circle while $\theta$ ranges between 0 and $\pi/4$, and the other where r ranges from 0 to the line y = 1 while $\theta$ ranges between $\pi/4$ and $\pi/2$.

Probably the hardest part of this problem is figuring out what the polar limits of integration are.