# How would I use this substitution for this volume integral?

I'm stuck on the following question, and would like some guidance or solution to this:
Let $V$ be the solid bounded by $z=2x$ above and below by $z=x^2 + y^2$.
First it tells me to express the volume $V$ in cylindrical coordinates, and I hope this is correct, please correct me if I'm wrong:
$$V = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{2\cos\theta}^{2}\int_{r^2}^{2r\cos\theta}r \mathrm{d}z\mathrm{d}r\mathrm{d}\theta.$$

After this, it then tells me to employ the substitution:
$$x=1+u\cos v, \quad y=u\sin v, \quad z=w.$$

I found the determinant of the jacobian to be $u$, but I have no clue how to change the boundaries with the substitutions.

The integral in cartesian coordinates is (I derived this, I hope this is correct):

$$V=\int_{0}^{2}\int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}}\int_{x^2+y^2}^{2x}\mathrm{d}z\mathrm{d}y\mathrm{d}x$$ for clarity.

How would I change my volume integral using their substitutions?

• The Cartesian bounds look fine to me. I think the integration over $r$ should run from $0$ to $2\cos\theta.$ – David K Jun 9 '17 at 11:47
• Thank you, and the $\theta$ bounds are fine right (I just changed it)? I don't ever recall having negative bounds for $\theta$ in polar substitutions, but it seems that it's necessary here. – Twenty-six colours Jun 9 '17 at 11:48
• The angle bounds look right. People often do zero to $2\pi$ for a full circle but the negative angle seems a lot more convenient here. – David K Jun 9 '17 at 11:52

The intersection of the paraboloid $z=x^2+y^2$ and the plane $z=2x$ is given by an ellipse, whose projection on the $xy$-plane is given by the conic with equation $x^2-2x+y^2=0$, that is a circle with unit radius centered at $(1,0)$. By setting $x=1+\rho\cos\theta$ and $y=\rho\cos\theta$ we have that on the given solid $z$ is bounded between $2+2\rho\cos\theta$ and $1+\rho^2+2\rho\cos\theta$, hence the wanted volume is given by
$$\int_{0}^{2\pi}\int_{0}^{1}\int_{1+\rho^2+2\rho\cos\theta}^{2+2\rho\cos\theta}\rho\,dz\,d\rho\,d\theta = 2\pi\int_{0}^{1}\rho(1-\rho^2)\,d\rho = \color{red}{\frac{\pi}{2}}.$$
• Ahhhh I see. So their given substitution is basically a "polar" substitution, except it """shifts""" our point of view of the origin, to the centre of the circle in the $x-y$ plane (so basically it's like we evaluate a triple integral, except we pretend that the projection was a circle centred at the origin)? – Twenty-six colours Jun 9 '17 at 11:45