Calculate the volume between these surfaces...

Question

In this exercise, I need to calculate the volume between these surfaces: Let $R>0$:

1. The cylinder: $x^2+y^2-2Ry=0$
2. The paraboloid: $z=2R^2-x^2-y^2$
3. The plain: $z=0$

I'm stuck because the cylinder is centered in $(x,y)=(0,R)$ and the paraboloid is centered in $(x,y)=(0,0)$. For the integral, it must be $z \in (0, 2R^2)$, which is the top of the paraboloid.

Actually, I only need to know hot to calculate the area, for every $z_0$, between the circumferences $x^2+^2-2Ry=0$ and $2R^2-z_0-x^2-y^2=0$, but the integrals without change of variables look terrible. Any ideas?

Thanks!

Answer 1

If you sketch the surfaces you will see that the hard part is to work out the region in the $z=0$ plane formed by the intersection of $z=0$ cross-section of the cylinder and the $z=0$ cross-section of the paraboloid. The height above each $(x,y)$ in this intersection is easily seen to be given by the paraboloid equation $z=2R^2-(x^2+y^2)$. Then we just integrate.

In $z=0$ the cylinder cross-section is just $x^2+y^2-2Ry=0$. Completing the square gives $x^2+(y-R)^2=R^2$ so it is a circle of radius $R$ with center $(0,R)$.

In $z=0$ the paraboloid cross-section is just $0=2R^2-(x^2+y^2)$ or $x^2+y^2=2R^2$ which is a circle of radius $\sqrt{2}R$ and center $(0,0)$.

If you sketch these two cross-sections you will see that we need to find where their boundaries intersect. The diagram should make it clear that the intersection points are $(R,R)$ and $(-R,R)$.

If you can't do it by inspection then it can be done analytically:

First, solve the equation for the cylinder cross-section for $y$ in terms of $x$:

\begin{eqnarray*} x^2+(y-R)^2&=&R^2\\ (y-R)^2 &=&R^2-x^2\\ y-R &=&\pm\sqrt{R^2-x^2}\\ y &=& R\pm\sqrt{R^2-x^2} \end{eqnarray*}

Second, solve the equation for the paraboloid cross-section for $y$ in terms of $x$:

\begin{eqnarray*} x^2+y^2&=&2R^2\\ y^2 &=&2R^2-x^2\\ y &=&\pm\sqrt{2R^2-x^2} \end{eqnarray*}

Third, recognize that the region we are interested in is bounded by the lower half of the cylinder cross-section graph and the upper half of the paraboloid cross-section graph. Thus, equate the appropriate $y$ values and solve for $x$:

\begin{eqnarray*} R-\sqrt{R^2-x^2} &=& \sqrt{2R^2-x^2} \\ (R-\sqrt{R^2-x^2})^2 &=& 2R^2-x^2 \\ R^2-2R\sqrt{R^2-x^2} + R^2-x^2 &=& 2R^2-x^2\\ -2R\sqrt{R^2-x^2} &=& 0\\ \end{eqnarray*} Hence, $x=\pm R$. Plugging these $x$-values into, say, $y=R-\sqrt{R^2-x^2}$ gives $y=R$.

Now, we have all the parts to assemble the integral: \begin{eqnarray*} I=\int_{x=-R}^{x=R}\int_{y=R-\sqrt{R^2-x^2}}^{y=\sqrt{2R^2-x^2}} \int_{z=0}^{z=2R^2-(x^2+y^2)}1dzdydx &=& 2\int_{x=0}^{x=R}\int_{y=R-\sqrt{R^2-x^2}}^{y=\sqrt{2R^2-x^2}} \int_{z=0}^{z=2R^2-(x^2+y^2)}1dzdydx \end{eqnarray*} where we use symmetry to restrict the integration of $x$ from $[-R,R]$ to $[0,R]$ and double the value we get.

This integral is screaming out to be transformed to cylindrical coordinates: $(x,y,z)=(r\cos\theta,r\sin\theta,z)$ for which the Jacobian is $r$ so that $dx\cdot dy\cdot dz=r\cdot dr\cdot d\theta\cdot dz$.

The region in $z=0$ can be divided into two angular ranges due to the two curves bounding it: $\theta\in [0,\frac{\pi}{4}]$ and $\theta\in [\frac{\pi}{4},\frac{\pi}{2}]$.

For $\theta\in [0,\frac{\pi}{4}]$ we use $x^2+(y-R)^2 \leq R^2$ to work out the range of $r$:

\begin{eqnarray*} x^2+(y-R)^2 &\leq &R^2\\ x^2+y^2-2yR+R^2 &\leq &R^2\\ x^2+y^2-2yR &\leq &0\\ r^2-2yR &\leq &0\\ r^2&\leq &2yR\\ r^2&\leq &2rR\sin\theta\\ r&\leq &2R\sin\theta\\ \end{eqnarray*}

For $\theta\in [\frac{\pi}{4},\frac{\pi}{2}]$ we use $x^2+y^2 \leq 2R^2$ to work out that the range of $r$ is $0\leq r \leq \sqrt{2}R$.

Lastly we see that $0 \leq z \leq 2R^2-(x^2+y^2) = 2R^2-r^2$.

Now we can put the integral into cylindrical coordinates:

\begin{eqnarray*} I &=& 2\int_{\theta=0}^{\theta=\frac{\pi}{4}}\int_{r=0}^{r=2R\sin\theta}\int_{z=0}^{z=2R^2-r^2}1rdzdrd\theta + 2\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}\int_{r=0}^{r=\sqrt{2}R}\int_{z=0}^{z=2R^2-r^2}1rdzdrd\theta\\ &=& 2\int_{\theta=0}^{\theta=\frac{\pi}{4}}\int_{r=0}^{r=2R\sin\theta}(2R^2-r^2)rdrd\theta + 2\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}\int_{r=0}^{r=\sqrt{2}R}(2R^2-r^2)rdrd\theta\\ &=& 2\int_{\theta=0}^{\theta=\frac{\pi}{4}}\int_{r=0}^{r=2R\sin\theta}(2R^2r-r^3)drd\theta + 2\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}\int_{r=0}^{r=\sqrt{2}R}(2R^2r-r^3)drd\theta\\ &=& 2\int_{\theta=0}^{\theta=\frac{\pi}{4}}(R^2r^2-\frac{1}{4}r^4)|_{r=0}^{r=2R\sin\theta} d\theta + 2\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}(R^2r^2-\frac{1}{4}r^4)|_{r=0}^{r=\sqrt{2}R}d\theta\\ &=& 2\int_{\theta=0}^{\theta=\frac{\pi}{4}}(4R^4\sin^2\theta-4R^4\sin^4\theta) d\theta + 2\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}(2R^4-R^4)d\theta\\ &=& 8R^4\int_{\theta=0}^{\theta=\frac{\pi}{4}}(\sin^2\theta-\sin^4\theta)d\theta + 2R^4\int_{\theta=\frac{\pi}{4}}^{\theta=\frac{\pi}{2}}d\theta\\ &=& 8R^4\int_{\theta=0}^{\theta=\frac{\pi}{4}}(\sin^2\theta (1-\sin^2\theta))d\theta + R^4\frac{\pi}{2}\\ \end{eqnarray*}

Hopefully, there aren't too many mistakes in all this. And, I'll leave it to you to finish the integration using $\sin^2\theta=\frac{1}{2}[1-\cos 2\theta]$ and $\cos^2\theta=\frac{1}{2}[1+\cos 2\theta]$

Answer 2

The volume is better integrated in cylindrical coordinations. At $z=0$, the two surfaces are two circles $r=\sqrt2 R$ and $r=2R\sin\theta$, respectively. They intersect at $\theta =\frac\pi4, \>\frac{3\pi}4$. So, divide the integrating region into two sections as

$$V=2\int_0^{\frac\pi4}\int_0^{2R\sin\theta} ({2R^2-r^2} )rdr d\theta + 2\int_{\frac\pi4}^{\frac\pi2} \int_0^{\sqrt2R}(2R^2-r^2)rdr d\theta $$