Calculate $\int_{0}^{2}\left (\int_{0}^{\sqrt{2x-x^2}} \sqrt{x^2+y^2}dy\right )dx$
My work: I tried polar coordinates x=$\rho*\cos(\theta)$;y=$\rho*\sin(\theta)$
I know that $\theta\in[0,\frac{\pi}{2}]$ but how can i find $\rho$?
My guess is $\rho\in[0,\frac{1}{\cos(\theta)}]$ ? Am i wrong? And if soo can somebody explain it to me.
The bound you are interested in is $y=\sqrt{2x-x^2}$, how does that translate over the polar.
If we square we have $y^2=2x-x^2$ or $x^2+y^2=2x$. Then if we translate to polar this gives, $r^2=2r\cos (\theta)$ which means $r=2\cos(\theta)$ is the bound. So then, remembering the Jacobian, we have,
$$\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\cos(\theta)} r^2 dr d\theta$$
HINT with another approach, for the inner integral, substitute $\text{y}=x\tan\left(\text{u}\right)$:
$$\mathcal{I}\left(x\right)=\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+\text{y}^2}\space\text{d}\text{y}=x^2\int_0^{\arctan\left(\frac{\sqrt{2-x}}{\sqrt{x}}\right)}\sec^3\left(\text{u}\right)\space\text{d}\text{u}\tag1$$
Now, use the reduction formula:
$$\int\sec^\text{n}\left(\text{u}\right)\space\text{d}\text{u}=\frac{\sec^{\text{n}-2}\left(\text{u}\right)\tan\left(\text{u}\right)}{\text{n}-1}+\frac{\text{n}-2}{\text{n}-1}\int\sec^{\text{n}-1}\left(\text{u}\right)\space\text{d}\text{u}\tag2$$
So, we get:
$$\mathcal{I}\left(x\right)=\frac{x^2}{2}\cdot\left(\int_0^{\arctan\left(\frac{\sqrt{2-x}}{\sqrt{x}}\right)}\sec\left(\text{u}\right)\space\text{d}\text{u}+\sec\left(\text{u}\right)\tan\left(\text{u}\right)\Big{|}_0^{\arctan\left(\frac{\sqrt{2-x}}{\sqrt{x}}\right)}\right)\tag3$$
Now, use that:
$$\int\sec\left(\text{u}\right)\space\text{d}\text{u}=\ln\left|\tan\left(\text{u}\right)+\sec\left(\text{u}\right)\right|+\text{C}\tag4$$
So, we get that:
$$\mathcal{I}\left(x\right)=\frac{x^2}{2}\cdot\left(\ln\left|\frac{\sqrt{2}+\sqrt{2-x}}{\sqrt{x}}\right|+\frac{\sqrt{\frac{4}{x}-2}}{\sqrt{x}}\right)\tag5$$