Let $D=\big\lbrace (x,y)\in\mathbb{R}^2\mid\frac{1}{2}<x^2+y^2<3 ~\text{and}~ y>0\big\rbrace$
I want to compute $A$ the area of this domain.
$$A=\iint_{D}{x^2+y^2~dxdy}$$
By a change of variable to polar coordinates we get $\Delta=\big\lbrace (r,\theta)\mid\frac{1}{\sqrt{2}}<r<\sqrt{3} ~\text{and}~ 0< \theta < \pi \big \rbrace$ so
$$A=\int_{0}^{\pi}\int_{\frac{1}{\sqrt{2}}}^{\sqrt{3}}{(r^2\cos^2(\theta)+r^2\sin^2(\theta))r~drd\theta}$$
$$=\int_{0}^{\pi}\int_{\frac{1}{\sqrt{2}}}^{\sqrt{3}}{r^3~drd\theta}$$
$$=\frac{1}{4}\int_{0}^{\pi}{[r^4]^\sqrt{3}_{\frac{1}{\sqrt{2}}}~d\theta}=\frac{35}{16}\pi$$
My question is am I wrong about the domain $\Delta$ ? When I use basic geometry with circle areas I find
$$A=\frac{\pi}{2}\bigg(3-\frac{1}{2}\bigg)=\frac{5}{4}\pi$$
Or did I missed something else ?
Best Answer
You have the wrong integrand. For calculating area it should be $$\iint_D1\,dx\,dy$$