Find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$

Question

I'm trying to find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ but nothing works for me.

I made the plot and it looks like this:

How could you start? Any recommendation?

Answer 1

Try checking where the two surfaces intersect. Solve the given equations for $z$ and set them equal:

$$\begin{align*} 1-x^2-y^2&=x^2+(y-1)^2\\ 0&=2x^2+2y^2-2y\\ 0&=x^2+y^2-y\\ 0&=x^2+\left(y-\frac12\right)^2-\frac14\\ x^2+\left(y-\frac12\right)^2&=\frac1{2^2} \end{align*}$$

which corresponds to the cylinder of radius $\frac12$ with cross sections parallel to the $(x,y)$ plane and centered over the point $\left(0,\frac12,0\right)$.

With this in mind, a change to cylindrical coordinates is the "obvious" move. Take

$$\begin{cases}x=r\cos\theta\\y=\frac12+r\sin\theta\\z=z\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$$

The solid (call it $S$) is then given by the set

$$S=\left\{(r,\theta,z)\mid0\le r\le\frac12,0\le\theta\le2\pi,r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta\right\}$$

where the last inequality follows from

$$\begin{align*} &x^2+(y-1)^2\le z\le1-x^2-y^2\\ &\implies r^2\cos^2\theta+\left(r\sin\theta-\frac12\right)^2\le z\le1-r^2\cos^2\theta-\left(\frac12+r\sin\theta\right)^2\\ &\implies r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta \end{align*}$$

The volume is then

$$\iiint_S\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\frac12}\int_{r^2-r\sin\theta+\frac14}^{\frac34-r^2-r\sin\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$$

$$=\frac\pi{16}$$

Answer 2

Here is how I would do it.

1. Solve for $z$ in both and set equal to each other, to get the intersection curve in the $xy$-plane.
2. Integrate the difference between $z$'s over the inside of that curve to get $\Delta V$.

Answer 3

The height $h$ of the region at a given $x,\,y$ is$$h=1-2x^2-y^2-(y-1)^2=2(1/4-x^2-(y-1/2)^2),$$and the relevant $x,\,y$ are those for which this is $\ge0$. Parameterizing such $x,\,y$ by $x=\rho\cos\theta,\,y=\tfrac12+\rho\sin\theta$ with $0\le\rho\le\tfrac12,\,0\le\theta\le2\pi$ gives the volume element as$$hdxdy=h\rho d\rho d\theta=2(1/4-\rho^2)\rho d\rho d\theta,$$so the volume is$$4\pi\int_0^{1/2}\rho(1/4-\rho^2)d\rho=-\pi[(1/4-\rho^2)^2]_0^{1/2}=\frac{\pi}{16},$$in agreement with @user170231's answer in terms of a triple integral.