(Iterated Integrals) Volume between a Cone and a Sphere

[MrMaterial]

Homework Statement

This is a book problem, as follows: Find the volume between the cone x = \sqrt{y^{2}+x^{2}} and the sphere x^{2}+y^{2}+z^{2} = 4

Homework Equations

spherical coordinates:
p^{2}=x^{2}+y^{2}+z^{2}
\phi = angle from Z axis (as I understand it)
\theta = angle from x or y axis

The Attempt at a Solution

I am confused on what solid they are referring to, so I will do it in two different ways.

I will first solve it assuming it is the volume of the cone (cone with spherical top)



It seems obvious that this is best solved using spherical coordinates.

First looking at the cone, it is a cone that appears on its side as it travels in the direction of the x axis.
Looking at the sphere, it is simply a sphere centered at the origin with a radius of \sqrt{4}


At this point I start setting up the triple integral.

\int\int\intp^{2}sin(\phi)dpd\phid\theta

p is going to go from 0 to \sqrt{4}
\phi is going to go from \pi/4 to 3\pi/4
θ is going from -\pi/4 to \pi/4

Then i punch it into the calculator and I get 4\pi(\sqrt{2}/3)

The book answer is 16\pi((\sqrt{2}-1)/(3\sqrt{2}))

For my second answer I will assume they want me to find the volume of the sphere, then subtract my former answer (the cone with spherical top)

First I will find the volume of the sphere

p is going to go from 0 to \sqrt{4}
\phi is going to go from 0 to \pi
θ is going from 0 to 2\pi

punch it into the ol' calculator and I get 32\pi/3

then I subtract my previous answer and I get

27.59

The book answer to this problem in decimal form is 9.815


So I must have found the volume of the wrong solid, or i set up my integrals wrong? Can any of you tell by looking at my work?

thanks!

 

[tiny-tim]

Hi MrMaterial!

\int\int\intp^{2}sin(\phi)dpd\phid\theta

p is going to go from 0 to \sqrt{4}
\phi is going to go from \pi/4 to 3\pi/4
θ is going from -\pi/4 to \pi/4
…
So I must have found the volume of the wrong solid, or i set up my integrals wrong?

(btw, you do know what √4 is don't you? )

(i assume you mean x = √(y² + z²))

i] You've missed a fundamental feature of limits in multiple integrals …

only the first limits are always between fixed values,

the next limits usually depend on the value of the first variable

ii] Also, in this case wouldn't it be a lot easier if you took your x axis as φ = 0 instead of the usual z axis ?

 

[HallsofIvy]

For x, y, z (rectangular) coordinates the limits on all three integrals are constants only if the region is a rectangular solid. For cylindrical coordinates the limits on all three integrals are constants only if the region is a cylinder. For spherical coordinates, the limits on all three integrals are constants only if the region is a sphere.

 

[MrMaterial]

ok so it turns out that i was actually finding more of a square cone (piece of a sphere) rather than an actual cone with a spherical top.

Now taking the comments into consideration i will find where both functions intersect.

After substituting and whatnot I get that the sphere and the cone intersect at 2 = z^{2}+y^{2}



Now i am going to reset the bounds on my dpd\phid\theta integral.

well I converted x = \sqrt{y^{2}+z^{2}} to spherical coordinates and solved for p and here's my guess as to what that bound will look like:
p goes from \sqrt{(1-tan(\theta)^{2})/(cos(\phi)^{2})} to 2

I couldn't seem to find \phi in terms of \theta so i set the remaining bounds to

\phi goes from \pi/4 to 3\pi/4
\theta goes from \pi/4 to -\pi/4

I also decided to do cartesian coordinates at one point!

for an integral of dxdydz I have the bounds:

x goes from \sqrt{z^{2}+y^{2}} to \sqrt{4-y^{2}-z^{2}}
y goes from -\sqrt{2-z^{2}} to \sqrt{2-z^{2}}
and z goes from 0 to \sqrt{2}

do any of these look right? I would evaluate them but my classpad isn't up to the task and I don't feel like doing all the pencil work if they aren't going to be right!