Equation in spherical and cylindrical coordinates

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numberonenacho

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Nov 19, 2007
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A) Write the equation of "rho=2sin phi" in rectangular and spherical coordinates.
B) Find the volume bounded by the surface rho=2sin phi and the xy plane, using triple integral in spherical coordinates
C) Use integration to locate the centroid of the volume described in b)

Any tips, help, work or similar problems would be nice.
Thanks~!
 
numberonenacho said:
A) Write the equation of "rho=2sin phi" in rectangular and spherical coordinates.
B) Find the volume bounded by the surface rho=2sin phi and the xy plane, using triple integral in spherical coordinates
C) Use integration to locate the centroid of the volume described in b)

Any tips, help, work or similar problems would be nice.
Thanks~!
Click to expand...

Looks like a take-home exam!

Please show us what you have tried - and exactly where you are stuck.
 
So for rectangular coordinates Im supposed to use the formulas x = rho sin(phi) * cos(theta), y = rho sin(phi) * sin(theta), and
z=rho cos(phi) & rho^2 = x^2 + y^2 + z^2 right?
I dont see how I could substitute those in to make the equation in rectangular coordinates.
So would I square both sides to make it rho^2 = 4sin(theta)^2
and then use the formula making it x^2 + y^2 + z^2 = 4sin(theta)^2
from there I have no idea how to get rid of the theta, or if im on the right track. >_<
 
Try to visualize - what does the surface look like?

If I were to describe it - I would say it looks like a sphere - symmetric about z-axis (circular symmetry - no theta dependence) - whose surfaces are defined by by a circle with center at z = 1 and radius = 1 (rotated around z-axis to make the sphere).

This is enough hint for a take-home exam - now show us some imagination and work.

I'll try to remember to put the solution next week.
 
given equation:

r = 2 * sin (phi)..............................(1)

sin(phi) = sqrt(x^2 + y^2)/r..............(2)

r = sqrt(x^2 + y^2 + z^2)

then, using (1) & (2)

r^2 = 2 * sqrt(x^2 + y^2)

Using (3)

x^2 + y^2 + z^2 = 2 * sqrt(x^2 + y^2) .................done ...part (A)
 
I believe your triple integral will look something like this(spherical coordinates).

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{0}^{{\rho}sin({\phi})}{\rho}^{2}sin({\phi})d{\rho}d{\phi}d{\theta}\)
 
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