Halfsphere, integral.. Please help!

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OrangeOne

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Sep 8, 2010
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The body K is described by the halfsphere:
x^2+y^2+z^2 equals to or is less than 1,
y is bigger than or equal to 0

I have these integrals:

??? x dxdydz
??? y dxdydz
??? z dxdydz
??? (x^2- z^2) dxdydz

The prof told me some of these can be determined directly by symmetric motivations, without doing any actual calculations. I don't understand this at all, can someone please explain this to me?

He also wants us to solve all the integrals.

By introducing spherical coordinates I get:

x= rsin?cos?
y= rsin?sin?
z= rcos?

r goes from 0 to 1?? Is this correct for my area?
? goes from 0 to 2pi Is this correct?
? goes from...???? (usually from 0 to pi, but we have one halfsphere..so what should it be in that case?)


I would really appreciate some help! Thanks.
 
In spherical, the volume of the hemisphere of radius 1 is:

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}{\rho}^{2}sin{\phi} \;\ d{\rho}d{\phi}d{\theta}\)
 
Thank you, but I thought that was the integral for when z is larger than (or equal to) 0.
When y is larger than or equal to zero, shouldn't it be that ? goes from -pi to pi, and ? goes from 0 to pi?
 
When y is larger than or equal to zero, shouldn't it be that ? goes from -pi to pi, and ? goes from 0 to pi?
 
What I showed was the triple integral for a hemisphere of radius 1 centered at the origin.

A hemisphere of radius 1 has volume \(\displaystyle \frac{2}{3}{\pi}\).

Because of symmetry, \(\displaystyle \int_{0}^{\pi}\int_{-\pi}^{\pi}\int_{0}^{1}{\rho}^{2}sin{\phi}d{\rho}d{\phi}d\theta=0\)

The way this is set up, it equals 0.

Using your first integral in rectangular, \(\displaystyle 2\int_{0}^{1}\int_{0}^{\sqrt{1-z^{2}}}\int_{0}^{\sqrt{1-z^{2}-y^{2}}}x \;\ dxdydz=\frac{\pi}{8}\)
 
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