Answer to Question #139322 in Calculus for Promise Omiponle

Question

Evaluate the following integral after converting to polar coordinates: (0 to 1) ∫ (y to sqrt(2-y^2)) ∫ (x+y) dx dy

Answer 1

Area of integration is a sector $\frac{\pi}{4}$ of a circle with radius $\sqrt{2}$ . For calculation in polar coordinates use formula

$\iint_{D}f(x,y)dxdy=\iint_{G}f(r\cos{\phi},r\sin{\phi})rdrd\phi$

$\int_{0}^{1}\int_{y}^{\sqrt{2-y^2}}(x+y)dxdy=$ (in the first answer here was a little error)

$=\int_{0}^{\sqrt{2}}\int_{0}^{\frac{\pi}{4}}(r\cos{\phi}+r\sin{\phi})rd\phi dr=$

$=\int_{0}^{\sqrt{2}}\int_{0}^{\frac{\pi}{4}}(\cos{\phi}+\sin{\phi})d\phi r^2dr=$

$=\int_{0}^{\sqrt{2}}(\sin{\phi}-\cos{\phi})|_{0}^{\frac{\pi}{4}}r^2dr=$

$=\int_{0}^{\sqrt{2}}r^2dr=\frac{1}{3}r^3|_{0}^{\sqrt{2}}=\frac{2\sqrt{2}}{3}$