Evaluate-R-xy-1-y-2-dx-dy-where-the-region-of-integration-is-the-positive-quadrant-of-the-circle-x-2-y-2-1-

Question Number 92805 by niroj last updated on 09/May/20

Evaluate :

\int_{R} \int \frac{xy}{\sqrt{1 - y^{2}}} dx dy where the region of integration is the

positive quadrant of the circle x^{2} + y^{2} = 1 .

Answered by mr W last updated on 09/May/20

= \int_{0}^{\frac{π}{2}} \int_{0}^{1} \frac{r^{2} \cos θ \sin θ}{\sqrt{1 - r^{2} \sin^{2} θ}} r d r d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos θ \sin θ (\int_{0}^{1} \frac{r^{2}}{\sqrt{1 - r^{2} \sin^{2} θ}} {d r}^{2}) d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos θ \sin θ (\int_{0}^{1} \frac{u}{\sqrt{1 - u \sin^{2} θ}} d u) d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos θ \sin θ \frac{1}{\sin^{2} θ} {\int_{0}^{1} [- \sqrt{1 - u \sin^{2} θ} + \frac{1}{\sqrt{1 - u \sin^{2} θ}}] d u} d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ} {{[\frac{2}{3 \sin^{2} θ} {(1 - u \sin^{2} θ)}^{\frac{3}{2}} - \frac{2}{\sin^{2} θ} \sqrt{1 - u \sin^{2} θ}]}_{0}^{1}} d θ

= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ} {[\frac{2}{3 \sin^{2} θ} {(1 - \sin^{2} θ)}^{\frac{3}{2}} - \frac{2}{3 \sin^{2} θ} - \frac{2}{\sin^{2} θ} \sqrt{1 - \sin^{2} θ} + \frac{2}{\sin^{2} θ}]} d θ

= \frac{1}{3} \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ} [\frac{\cos^{3} θ - 3 \cos θ + 2}{\sin^{2} θ}] d θ

= \frac{1}{3} \int_{\frac{π}{2}}^{0} \frac{(\cos^{3} θ - 3 \cos θ + 2) \cos θ}{{(1 - \cos^{2} θ)}^{2}} d (\cos θ)

= \frac{1}{3} \int_{0}^{1} \frac{(t^{3} - 3 t + 2) t}{{(1 - t^{2})}^{2}} d t

= \frac{1}{3} \int_{0}^{1} \frac{(t + 2) t}{{(t + 1)}^{2}} d t

= \frac{1}{3} \int_{0}^{1} [1 - \frac{1}{{(t + 1)}^{2}}] d t

= \frac{1}{3} {[t + \frac{1}{t + 1}]}_{0}^{1}

= \frac{1}{3} (1 + \frac{1}{2} - 1)

= \frac{1}{6}

Commented by niroj last updated on 09/May/20

Thank you Mr. W It is aspected result ��

Commented by Ar Brandon last updated on 09/May/20

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