How Do You Calculate the Total Mass of a Cylinder with Variable Density?

TheSpaceGuy
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Total mass?

Homework Statement


Find the total mass of the part of the solid cylinder x^2 + y^2 ≤ 4 such that x^2 ≤ z ≤ 9 - x^2 , assuming that the mass density is p(x, y, z) = I y I (absolute value of y).


I have heard about center of mass but this is throwing me off?

The Attempt at a Solution



Thats where the problem is.
 
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All you have to do is integrate p(x,y,z) over the volume of interest.
 


But how would I get the limits of integration. How about choosing x from 0 to 4 and y is x^2 to 4? z is given. Am I on the right track?
 


Did you think about this very long? x can't 4 and satisfy x^2+ y^2= 4 for any y! And I have no idea how you got "y is x^2 to 4"! What do you get if you solve x^2+ y^2= 4 for y?

Perpendicular to the z-axis, the boundary is the cylinder x^2+ y^2= 4. You could let x very from -2 to 2 and, then, for every x, y varies from -\sqrt{4- x^2} to \sqrt{4- x^2}. Or write it in polar coordinates with r going from 0 to 2, \theta from 0 to 2\pi.

For every point (x, y), the z-coordinate varies from x^2 to 9- x^2 just as you are told.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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