Sphere using mulitiple integration

[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

 

[HallsofIvy]

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

My first reaction when I read this was "find a what??" 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.

 

[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.

 

[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!

 

[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 realized the answer to as soon as i clicked on submit reply lol

thanks for the help

 

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

 

[HallsofIvy]

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

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.)

 

[AlmostFamous]

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

 

[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.

 

[HallsofIvy]

No, the limits of integration, after the substitution, are no longer 0 and a. Also, the anti-derivative of u^1/2 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.

 

[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

 

[HallsofIvy]

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

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