Evaluate the iterated integral by converting to polar coordinates, ??(x²+y²)dydx

Question

Calculus 3: Double Integral

Conditions are that, the first integral ranges from (0→2). 

The second integral ranges from (0→√4-x²). the square root covers the whole 4-x² and not just 4 itself.

Answer 1

We know that:

x=rcosθ and y=rsinθ with 0≤r≤2 and 0≤θ≤pi.

Also we need the Jacobian Determinant ∂(x,y)/∂(r,θ) = r (in this case). 

Therefore the double integral becomes int_{θ=0→pi}int_{r=0→2}(rcosθ)^2+(rsinθ)^2 r drdθ. 

This turns out to be int_{θ=0→pi}int_{r=0→2}r^3drdθ=pi/4(2)^4=4pi.

 

I hope you understand the typing.