Question Number 78705 by abdomathmax last updated on 20/Jan/20
$${calculate}\:\int\int_{{D}} \:\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){dxdy}\: \\ $$$${D}\:=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant\mathrm{3}\:{and}\:{y}\geqslant\mathrm{0}\right\} \\ $$
Answered by mind is power last updated on 20/Jan/20
![x=rcos(θ),y=rsin(θ)⇒dxdy=rdrdθ D⇔(1/2)≤r^2 ≤3 ^ sin(θ)≥0⇒θ∈[0,π] r∈[(1/( (√2))),(√3)] ∫∫_D (x^2 +y^2 )dxdy=∫_0 ^π ∫_(1/( (√2))) ^(√3) r^3 drdθ=π.[(r^4 /4)]_(1/( (√2))) ^(√3) =π((9/4)−(1/(16)))=((35π)/(16))](https://www.tinkutara.com/question/Q78737.png)
$$\mathrm{x}=\mathrm{rcos}\left(\theta\right),\mathrm{y}=\mathrm{rsin}\left(\theta\right)\Rightarrow\mathrm{dxdy}=\mathrm{rdrd}\theta \\ $$$$\mathrm{D}\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{r}^{\mathrm{2}} \leqslant\mathrm{3}\:\bar {\:}\mathrm{sin}\left(\theta\right)\geqslant\mathrm{0}\Rightarrow\theta\in\left[\mathrm{0},\pi\right] \\ $$$$\mathrm{r}\in\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\sqrt{\mathrm{3}}\right] \\ $$$$\int\int_{\mathrm{D}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy}=\int_{\mathrm{0}} ^{\pi} \int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\sqrt{\mathrm{3}}} \mathrm{r}^{\mathrm{3}} \mathrm{drd}\theta=\pi.\left[\frac{\mathrm{r}^{\mathrm{4}} }{\mathrm{4}}\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\sqrt{\mathrm{3}}} \: \\ $$$$=\pi\left(\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{16}}\right)=\frac{\mathrm{35}\pi}{\mathrm{16}} \\ $$