# Cartesian to polar

## Homework Statement

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.
int(-1to1)int((sqrt(1-y^2))to(sqrt(1-y))[x^2+y^2]dxdy

x=rcostheta
y=rsintheta

## The Attempt at a Solution

int(-1to1)int((sqrt1-(rsintheta)^2)to(sqrt(1-(rsintheta))))[(rcostheta)^2+(rsintheta)^2]rdrdtheta

CompuChip
Homework Helper
Note that [x^2+y^2] is simply r^2 (by definition!).

Also you cannot just change your integration boundaries like that. For example, neither theta nor r runs between -1 and 1.
I suggest drawing an image of the integration region, then think about how to describe it in terms of boundaries on r and theta.

ok so i am not sure how to get the new boundries
but its r^2rdrdtheta which integrated goes to r^4/4dtheta. Is this right?

HallsofIvy
Yes, $\int r^3dr$ is $r^4/4$.
Are you sure about that "$y= \sqrt{1- x}$ limit? That will give the right part of a parabola for y> 0 it is inside the unit circle given by the lower limit and for y< 0, it is outside.