# Thread: Volume of a Solid

1. ## Volume of a Solid

Find the volume of a solid that lies above the plane $xOy$ and is bounded by the surfaces $x^2+y^2+z^2=1$ and $x^2+y^2-z^2=0$.

Start of a solution:
We can find $f(x,y) = z$, i.e., $z = \sqrt{1-x^2-y^2}$ and $z = \sqrt{x^2+y^2}$.

I know that this involves multiple integrals, but my book was not very helpful and I cannot remember how to set this up.

2. ## Re: Volume of a Solid

Hey vidomagru.

Whenever multiple integrals come, the first thing you want to do is check what kind of co-ordinate system comes up.

Hint: Take a look at the standard co-ordinate systems and see which one is best suited (take a look at cartesian, cylindrical, polar, and paraboloid systems).

3. ## Re: Volume of a Solid

Originally Posted by chiro
Hey vidomagru.

Whenever multiple integrals come, the first thing you want to do is check what kind of co-ordinate system comes up.

Hint: Take a look at the standard co-ordinate systems and see which one is best suited (take a look at cartesian, cylindrical, polar, and paraboloid systems).
Ok so I believe that the best system to use would be cylindrical. Here is what I have done so far:

we have $x^2+y^2+z^2=1$ with $x=rcos\theta$ and $y=rsin\theta$. So substituting these in we get
$(rcos\theta)^2 + (rsin\theta)^2 + z^2 = 1$

$r^2(cos^2\theta + sin^2\theta) + z^2 = 1$

$r^2 + z^2 = 1$

$z = \pm \sqrt{1-r^2}$, so the limits for $z$ are $(-\sqrt{1-r^2},\sqrt{1-r^2})$.

Now for $x^2+y^2-z^2=0$ we have:
$(rcos\theta)^2 + (rsin\theta)^2 - z^2 = 0$

$r^2(cos^2\theta + sin^2\theta) = z^2$

$r^2 = z^2$

$r = \pm z$, so the limits for $r$ are $(-z,z)$.

And the limits for $\theta$ are $(0,2\pi)$.

So now we can set up the triple integral to be $\int \int \int f(x,y,z)dV = \int_0^{2\pi} \int_{-z}^z \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r dzdrd\theta$. So let's write this as:

$\int_0^{2\pi} \int_{-z}^z \left( \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r dz \right) drd\theta$

$=\int_0^{2\pi} \int_{-z}^z \left[rz\right]_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} drd\theta$

$=\int_0^{2\pi} \int_{-z}^z \left(r(\sqrt{1-r^2}) - r(-\sqrt{1-r^2})\right) drd\theta$

$=\int_0^{2\pi} \int_{-z}^z \left(2r(\sqrt{1-r^2})\right) drd\theta$

$=\int_0^{2\pi} \left[\frac{-2}{3}(1-r^2)^\frac{3}{2}\right]_{-z}^z d\theta$....

So I would continue here, but I have obviously made a mistake since I do not want this to be in terms of $z$... Help?

4. ## Re: Volume of a Solid

Still confused on this, any thoughts?

5. ## Re: Volume of a Solid

The surfaces are given by $x^2+ y^2+ z^2= r^2+ z^2= 1$ and $x^2+ y^2- z^2= r^2- z^2= 0$ or $z^2= r^2$. Where they intersect, $r^2+ r^2= 2r^2= 1$ so that $r= 1/\sqrt{2}= \sqrt{2}/2$. r goes from 0 to $\sqrt{2}/2$. Then for each r z goes from $r$ up to $\sqrt{1- r^2}$.

6. ## Re: Volume of a Solid

Originally Posted by HallsofIvy
The surfaces are given by $x^2+ y^2+ z^2= r^2+ z^2= 1$ and $x^2+ y^2- z^2= r^2- z^2= 0$ or $z^2= r^2$. Where they intersect, $r^2+ r^2= 2r^2= 1$ so that $r= 1/\sqrt{2}= \sqrt{2}/2$. r goes from 0 to $\sqrt{2}/2$. Then for each r z goes from $r$ up to $\sqrt{1- r^2}$.

I might just be missing something obvious here, but why does r not go from $-\sqrt{2}/2$ and instead begin at zero? And why does z start from $r$?

7. ## Re: Volume of a Solid

In cylindrical coordinates, 0 is the smallest value r can take. Then you have $z^2 = r^2$ and $z^2 = 1-r^2$. One of them is the lower bound and one is the upper bound. Since you are integrating over $\theta$ from 0 to $2\pi$, you are picking up all of the negatives.