Triple integral from cartesian to spherical coordinates

[smashyash]

Homework Statement

evaluate the following triple integral in spherical coordinates::

INT(=_B) = (x^2+y^2+z^2)^2 dz dy dx

where the limits are:

z = 0 to z = sqrt(1-x^2-y^2)
y = 0 to z = sqrt(1-x^2)
x = 0 to x = 1

Homework Equations

The only thing I know for sure is how to set up the spherical function:

INT_B = p^6 sin(phi) dp d(theta) d(phi)

The Attempt at a Solution

Now, how do you evaluate the limits??

I assume it's a cylinder so are the theta limits 0 to 2pi?

 

[Dick]

It looks like an octant of a sphere to me.

 

[smashyash]

It looks like an octant of a sphere to me.

Oh, yes. I'm mistaken...

Unfortunately, I still don't know how to finish setting up with new integral...

 

[I like Serena]

Oh, yes. I'm mistaken...

Unfortunately, I still don't know how to finish setting up with new integral...

You already have the integral. You just need the bounds.
Which bounds would correspond to the octant of a sphere?

 

[Dick]

This is about as easy as setting up limits in spherical coordinates get. Look up a nice picture of what the coordinates in spherical coordinates look like. Like here http://mathworld.wolfram.com/SphericalCoordinates.html Then picture your octant on top of that.

 

[smashyash]

This is about as easy as setting up limits in spherical coordinates get. Look up a nice picture of what the coordinates in spherical coordinates look like. Like here http://mathworld.wolfram.com/SphericalCoordinates.html Then picture your octant on top of that.

Thanks you! That's a great visual aid!

You already have the integral. You just need the bounds.
Which bounds would correspond to the octant of a sphere?

This is still the trouble I have... visualizing exactly where this sphere is or what it's boundaries are..

I would think that both theta and phi are from 0 to pi/2 and I'm really unsure about p..

 

[HallsofIvy]

z= \sqrt{1- x^2- y^2
is equivalent to saying that z is non-negative and that x^2+ y^2+ z^2= 1.
That's the part of the sphere, with center at (0, 0, 0) and radius 1, in the first octant.

Normally, \theta would go from 0 to 2\pi to cover the entire circle in the xy plane. Here, you only want to cover the first quadrant. How should \theta vary?

Normally, \phi would go from 0 to \pi to cover the entire half circle from the positive z-axis to the negative z-axis. Here, you only want to go half way, from the positive z-axis to z= 0. How should \phi vary?

Of course, \rho goes from 0 out to the sphere of radius 1.