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

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?

• One place to start might be to look at the intersection of the two surfaces. What does this curve look like when projected onto the $x$-$y$ plane (i.e. eliminate $z$ from the two equations and look at the resulting equation in $x$ and $y$). – John Don Jun 2 at 18:11

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

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

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

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.

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 $$\Delta V$$.