# Homework Help: Sphere using mulitiple integration

1. May 15, 2006

### AlmostFamous

I've proved the volume of a sphere using multiple integration but now i need to use that for the following:

find a so that the volume inside the hemisphere z=sqrt(16-x^2-y^2) and outside the cylinder $$x^2 + y^2= a^2$$ is one quarter of the hemisphere.

I'm having trouble even formulating an integral for this. Once it is set up as a multiple integral i'll be okay with it. Any help would be appreciated. Thanks

2. May 15, 2006

### HallsofIvy

My first reaction when I read this was "find a what??":rofl: you are to find a number a so that the volume inside the hemisphere and outside the cylinder is 1/4 the volume of the hemisphere. That is the same as saying the volume inside the hemisphere and inside the cylider is 3/4 the volume of the hemisphere. Of course, a sphere of radius 4 has volume $\frac{4}{3}\pi(4)^3= \frac{216}{3}\pi$ and the hemisphere has volume $\frac{128}{3}\pi$so you are looking for a that will make the volume 3/4 of that: $32\pi$.

I would recommend using cylindrical coordinates (obviously, a is less than 4). Then the integral with respect to r is from 0 to a (unknown), the integral with respect to $\theta$ is from 0 to $2\pi$, and the integral with respect to z is from 0 to the hemisphere: 0 to $\sqrt{16- r^2}$.

$$\int_0^{2\pi}\int_0^a\int_0^{\sqrt{16-r^2}}rdzdrd\theta= 2\pi\int_0^a\sqrt{16-r^2}rdr$$

That, of course, is a very simple integral which will give you the volume in terms of a. Set it equal to $32\pi$ and solve for a.

Last edited by a moderator: May 16, 2006
3. May 15, 2006

### AlmostFamous

shouldn't that be setting it equal to 32/pi not 32pi?

also where does the equation $\sqrt{16- r^2}$ come from?

is this possible to do in cartesian coords? or is this way just easier.

4. May 15, 2006

### HallsofIvy

Why in the world would it be 32/pi? The volume of a hemisphere of radius 4 is pi times(2/3)(43) and 3/4 of that is $32\pi$.
The equation you gave for the hemisphere was $z= \sqrt{16- x^2- y^}$. In polar coordinates (cylindrical coordinates without z) $x^2+ y^2= r^2$ so that becomes $z= \sqrt{16- r^2}$.

Yes, you can do it in Cartesian coordinates: the integral would be
$$\int_{x=-a}^a\int_{y=-\sqrt{a^2- x^2}}^{\sqrt{a^2-x^2}}\int_{z=0}^{\sqrt{16-x^2-y^2}}dzdydx= \int_{x=-a}^a\int_{y= -\sqrt{a^2- x^2}}^{\sqrt{16-x^2}}\sqrt{16- x^2- y^2}dydx$$
and you will have to do a rather complicated trig substitution to integrate that. I think you will find it is far simpler in polar coordinates!

5. May 15, 2006

### AlmostFamous

i thought it should be 32pi, but you'd got 32/pi further up in the question, so i was just checking.

and asking where the 16-r^2 part came from was a stupid question that i realised the answer to as soon as i clicked on submit reply lol

thanks for the help

6. May 16, 2006

### AlmostFamous

the final integral looks simple enough, but believe it or not...i just can't seem to do it. the more i look at it the more i can't do it.

$$2\pi\int_0^a\sqrt{16-r^2}rdr$$

could i get a hint as to how to do this? i'm guessing integration by parts, but i can't figure out how to integrate $$\int_0^a\sqrt{16-r^2}dr$$

7. May 16, 2006

### HallsofIvy

Try the substitution u= 16- r2!

($\int_0^a\sqrt{16-r^2}dr$ is a standard trig substitution: let $x= 4 sin(\theta)$ but having the r outside the integral makes $2\pi\int_0^a\sqrt{16-r^2}rdr$ much easier.)

8. May 16, 2006

### AlmostFamous

i was never particularly good with integration by substitution, but i'll give this a go. thanks

9. May 16, 2006

### AlmostFamous

okay, i've had a go at this and i end up with
$2pi\frac{(16-r^2)^2}{4}=32pi$ between the limits of a and 0, where a is the number i'm trying to find. then i'd go on to substitue a in and solve.

could someone tell me if this is right, or have i gone drastically wrong somewhere.

10. May 16, 2006

### HallsofIvy

No, the limits of integration, after the substitution, are no longer 0 and a. Also, the anti-derivative of u1/2 will involve a factor of 2/3 and I don't see that anywhere. If u= 16- r2 then du= -2r dr so -(1/2)du= rdr. Of course, $\sqrt{16-r^2}= u^{\frac{1}{2}}$. When r= 0, u= 16 and when r= a, u= 16-a2 so the integral becomes
$$\pi\int_{16}^{16-a^2}u^{\frac{1}{2}}du= \frac{2\pi}{3}(16-a^2)^{\frac{3}{2}}= 32\pi$$

That should be easy to solve for a.

11. Mar 10, 2007

### ashnicholls

But solving this

$$\int_0^a\sqrt{16-r^2}dr$$

is not the answer to this

$$2\pi\int_0^a\sqrt{16-r^2}rdr$$

because the r times root 16 - r squared, why have you ignored the r.

How do you incorporate that?

Cheers Ash

12. Mar 10, 2007

### HallsofIvy

Which post are you referring to? Both Almost Famous and I, in our last posts, were referring to
$$2\pi\int_0^a\sqrt{16-r^2}rdr$$