I'm stuck on evaluating the integral
$$
\iiint\limits_V z \, \mathrm{d}x\, \mathrm{d}y\, \mathrm{d}z,
$$
where $V$ is bounded by the cone $z=\sqrt{2x^2+2y^2}$ and the plane $z=y+1$.
I've drawn the intersection of the two surfaces, and the projection onto $xy$-plane is the ellipse $x^2 + \frac{(y-1)^2}{2}=1$.
I could use a hint as to how to approach evaluating the integral.
I've tried using elliptical coordinates with translation: $x=r\cos\varphi$, $y=\sqrt{2}r\sin\varphi+1$, and $z=z$, but I'm not sure how to go about setting boundaries if I do that because $z\in[ \sqrt{2x^2+2y^2},y+1]$.
Am I making a mistake or is this a good approach?
Best Answer
Given the set:
$$ \Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : \sqrt{2\,x^2+2\,y^2} \le z \le y+1 \right\} $$
it follows that:
$$ I := \iiint\limits_{\Omega} z\,\text{d}x\,\text{d}y\,\text{d}z = \iint\limits_D \frac{(y + 1)^2 - \left(2\,x^2+2\,y^2\right)}{2}\,\text{d}x\,\text{d}y $$
where:
$$ \small D := \left\{ (x,\,y) \in \mathbb{R}^2 : \sqrt{2\,x^2+2\,y^2} \le y+1 \right\} = \left\{ (x,\,y) \in \mathbb{R}^2 : \frac{(x - 0)^2}{(1)^2} + \frac{(y - 1)^2}{\left(\sqrt{2}\right)^2} \le 1 \right\}. $$
So, opting for the following coordinate transformation:
$$ \Phi : \begin{cases} x = 0 + 1\,\rho\,\cos\theta \\ y = 1 + \sqrt{2}\,\rho\,\sin\theta \end{cases} \; \; \; \; \; \; \text{with} \; (\rho,\,\theta) \in [0,\,1] \times [0,\,2\pi] $$
and Jacobian $J_{\Phi} = 1 \cdot \sqrt{2} \cdot \rho$, all this translates into the following calculation:
$$ \small I = \int_0^{2\pi} \int_0^1 \frac{\left(2+\sqrt{2}\,\rho\,\sin\theta\right)^2 - 2\,\left(0+1\,\rho\,\cos\theta\right)^2 - 2\,\left(1+\sqrt{2}\,\rho\,\sin\theta\right)^2}{2}\,\sqrt{2}\,\rho\,\text{d}\rho\,\text{d}\theta = \frac{\pi}{\sqrt{2}}\,. $$
Finally, wanting to verify the result obtained in Wolfram Mathematica: