The evaluation of the double integral is [tex]\mathbf{ \dfrac{105}{2}\pi }[/tex]
The double integral [tex]\mathbf{\int \int _R\ \dfrac{y^2}{x^2+y^2} \ dA}[/tex], where R is the region that lies between
the circles [tex]\mathbf{x^2 +y^2 = 16 \ and \ x^2 + y^2 = 121}[/tex].
Let consider x = rcosθ and y = rsinθ because x² + y² = r²;
Now, the double integral can be written in polar coordinates as:
[tex]\mathbf{\implies \int \int _R\ \dfrac{y^2}{x^2+y^2} \ dxdy}[/tex]
[tex]\mathbf{\implies \int \int _R\ \dfrac{r^2 \ sin^2 \theta}{r^2} \ rdrd\theta}[/tex]
[tex]\mathbf{\implies \int \int _R\ \ sin^2 \theta \ r \ drd\theta}[/tex]
Thus, the integral becomes:
[tex]\mathbf{=\int^{2 \pi}_{0} sin^2 \theta d\theta \int ^{11}_{4} rdr }[/tex]
∴
[tex]\mathbf{=\int^{2 \pi}_{0} \dfrac{1-cos 2 \theta }{2} \ \theta \ d\theta\dfrac{r}{2} \Big|^{11}_{4}dr }[/tex]
[tex]\mathbf{\implies \dfrac{1}{2} \Big[\theta - \dfrac{sin \ 2 \theta}{2}\Big]^{2 \pi}_{0} \ \times\Big[ \dfrac{11^2-4^2}{2}\Big]}[/tex]
[tex]\mathbf{\implies \dfrac{\pi}{2} \times\Big[ 121-16\Big]}[/tex]
[tex]\mathbf{\implies \dfrac{105}{2}\pi }[/tex]
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