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

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.
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1 Answer

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.