1. ## integral.

Does anyone know how to do this integral?
$\displaystyle \int_{x=0}^{x=3}\int_{y=0}^{y=\sqrt{9-x^2}}\sqrt{1+4x^2+4y^2}\,dy\,dx$ I think a substituition you can use is either sin or cos but i'm not too sure.

2. Originally Posted by oxrigby
Does anyone know how to do this integral?
$\displaystyle \int_{x=0}^{x=3}\int_{y=0}^{y=\sqrt{9-x^2}}\sqrt{1+4x^2+4y^2}\,dy\,dx$ I think a substituition you can use is either sin or cos but i'm not too sure.
Rewrite this as $\displaystyle \int_0^3\int_0^{\sqrt{9-x^2}}\sqrt{1+4\left(x^2+y^2\right)}\text{ }dy\text{ }dx$. Make a change of coordinates to polar.(don't forget the Jacobian) so that we get $\displaystyle \int_0^{\frac{\pi}{2}}\int_0^3 r\cdot\sqrt{1+4r^2}\text{ }dr\text{ }d\theta$

3. Originally Posted by oxrigby
Does anyone know how to do this integral?
$\displaystyle \int_{x=0}^{x=3}\int_{y=0}^{y=\sqrt{9-x^2}}\sqrt{1+4x^2+4y^2}\,dy\,dx$ I think a substituition you can use is either sin or cos but i'm not too sure.
Use polar coordinates!!!

$\displaystyle x=r\cos\theta$
$\displaystyle y=r\sin\theta$
$\displaystyle dA=r\,dr\,d\theta$

Here your area is the quarter circle in the first quadrant, so the bounds on $\displaystyle \theta$ are $\displaystyle 0..\frac{\pi}{2}$.

Your circle has radius $\displaystyle 3$, so the bounds on $\displaystyle r$ are $\displaystyle 0..3$.

So you're integrating

$\displaystyle \int_0^{\pi/2}\int_0^3 r\sqrt{1+4r^2}\,dr\,d\theta$

which is straightforward.