Converting Cartesian to Polar (Double Integral)

[Trebond]

Homework Statement

Integrate from 0 to 1 (outside) and y to sqrt(2-y^2) for the function 8(x+y) dx dy.
I am having difficulty finding the bounds for theta and r.

Homework Equations

I understand that somewhere here, I should be changing to
x = r cost
y = r sin t
I understand that I can solve for x^2 +y^2 = 2, so I believe that r should range from 0 to sqrt 2 unless I am mistaken.

The Attempt at a Solution

I understand that y = 0 to y = 1,
and x is from y to sqrt (2-y^2), but what are the bounds for x? Given that y could be either 0 or 1, does y sweep from 1 to sqrt (2-y^2) or 0 to sqrt(2-y^2)?
EDIT: I found the solution correctly, but I would like to understand polar more in depth. I seem to be struggling quite a lot. To solve, I believed that theta had to range from 0 to pi/4 because x is bounded by y = x and sqrt(2-y), and thus intersect when y and x = 1 (or theta = pi/4).

 

[BvU]

Hello Trebond,

If you draw a picture, the answer is evident !

 

[Mark44]

And conversely, if you don't draw a picture, a problem involving a change of coordinate systems is much more difficult.