Given, the integral ∫−20∫−4−y24−y2x2dxdy .
From the integral limit, the curves are
y=0 , ..........................................1
y=−2, ....................................2
x2+y2=4 . ..............................3
Convert cartesian to polar coordinate,
x=rcosθ, ................................4
y=rsinθ, .................................5
r=x2+y2, ........................6
θ=tan−1xy ..............................7
From equation 3 and 6,
r=2
Therefore, 0≤r≤2 and 0≤θ≤2π.
Then, the integral
∫−20∫−4−y24−y2x2dxdy=∫02π∫02r2cos2θ(rdrdθ)=∫02π∫02r3cos2θdrdθ=∫02π[4r4]02 cos2θdrdθ=∫02π1.cos2θdθ=∫02π21+cos2θdθ=21[θ+2sin2θ]02π=π
Thus, the value of the given integral is π.
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