Iterated integrals converted to polar

[mikky05v]

Homework Statement
∫^{4}_{0} ∫^{√(4y-y^{2})}_{0} (x²) dx dy


The attempt at a solution

I'm confused on how to convert the bounds into polar coordinates.

I believe x² just becomes r²cos²θ

0≤x≤√(4y-y²)
0≤y≤4

but i don't know how to convert the bounds

 

[Mark44]

Homework Statement
∫^{4}_{0} ∫^{√(4y-y^{2})}_{0} (x²) dx dy


The attempt at a solution

I'm confused on how to convert the bounds into polar coordinates.

I believe x² just becomes r²cos²θ

0≤x≤√(4y-y²)
0≤y≤4

but i don't know how to convert the bounds

Have you drawn a sketch of the region over which integration is taking place? That's very important when you're changing from one coordinate system to another.

 

[mikky05v]

ya I came up with a half circle in the 1st quadrant that's center is on 2 and it's radius is 2 but i can't figure out how to use that to find my bounds. If the center was on 0 I would be a lot more comfortable

 

[mikky05v]

I plugged in the values of x and y for polar coordinates and came up with r=4sinθ so would my r bounds simply go from 0→4sinθ?

 

[Mark44]

ya I came up with a half circle in the 1st quadrant that's center is on 2 and it's radius is 2 but i can't figure out how to use that to find my bounds. If the center was on 0 I would be a lot more comfortable

Your description isn't very clear, but I think you understand what the region looks like.

The equation of your circle is x² + y² - 4y = 0, where x ≥ 0 (for the right half).

Converting to polar form, we get r² - 4rsin(θ) = 0, or r(r - 4sin(θ)) = 0. Since there is a value of θ that is paired with r = 0, we don't lose any solutions by dividing by r.

The polar equation of your circle is r = 4sin(θ).