Question

Calcule a seguinte integral


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Answer 1

$\text{Se tiene la regi\'on }\\ R=\left\{(x,y): 0\leq x \leq a ; 0\leq y \leq \sqrt{a^2-x^2}\right\}\\ \\ \text{haremos:} \\ x=r\cos \theta \wedge y=r\sin \theta \\ \\ \text{donde: } |J(r,\theta)|=r \\ \text{por ende:}\\ R=\left\{(r,\theta): 0\leq r \leq a\; ;\; 0\leq \theta \leq \pi/2 \right\}\\ \\ \text{Entonces:}\\ \\ \displaystyle \int_{0}^{a}dx\int_{0}^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\;dy=\int_{0}^{a}\int_{0}^{\pi/2}r^2 dr\,d\theta$

$\displaystyle \int_{0}^{a}dx\int_{0}^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\;dy=\int_{0}^{a}dr\int_{0}^{\pi/2}r^2 \,d\theta\\ \\ \int_{0}^{a}dx\int_{0}^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\;dy=\frac{\pi}{2}\cdot\int_{0}^{a}r^2dr \\ \\ \int_{0}^{a}dx\int_{0}^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\;dy=\frac{\pi}{2}\left.\frac{r^3}{3}\right|_0^a\\ \\ \\ \boxed{\int_{0}^{a}dx\int_{0}^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\;dy=\frac{a^3\pi }{6}}$