1. ## Polar equivalent integral

Hi,

I'm supposed to change this cartesian integral into a polar integral.

Cartesian form:
(integral from y=0 to y=2)(integral from x=-(sqrt(1-(y-1)^2)) to x=0) of xy^2 dxdy

I got the polar equivalent to be:

(integral from theta=-pi/2 to pi/2)(integral from r=0 to r=1) of r^4(costheta)(sin^2theta) drdtheta

Have I done this correctly??

2. Originally Posted by s7b
Hi,

I'm supposed to change this cartesian integral into a polar integral.

Cartesian form:
(integral from y=0 to y=2)(integral from x=-(sqrt(1-(y-1)^2)) to x=0) of xy^2 dxdy

I got the polar equivalent to be:

(integral from theta=-pi/2 to pi/2)(integral from r=0 to r=1) of r^4(costheta)(sin^2theta) drdtheta

Have I done this correctly??
No. Your new integral terminals are wrong.

Note that the area is the left half of the circle $\displaystyle (y - 1)^2 + x^2 = 1$ so you need to express that circle in polar form.

Note that the left half of the circle is gven by $\displaystyle \frac{\pi}{2} \leq \theta \leq \pi$.

3. When I take the integral after changing the limits you said I end up with negative 1/15. Shouldn't the answer be positive??

4. Originally Posted by s7b
When I take the integral after changing the limits you said I end up with negative 1/15. Shouldn't the answer be positive??
Excuse me ....? Where did I say that??

Re: Expressing the circle in polar form. Reading this thread might help: http://www.mathhelpforum.com/math-he...variables.html

5. S7b, the integral must be negative. It is the integral of $\displaystyle xy^2$ and is taken over a region where x is always negative: x ranges from $\displaystyle -\sqrt{1- (y-1)^2}$ up to 0 for every y.