Register to reply 
Volume bounded above by cylinder z = 4  x^2, x^2 + y^2 = 4 on sides,xyplane bottom 
Share this thread: 
#1
Apr1305, 07:29 PM

P: 492

Here is the problem:
Find the volume of the solid that is bounded above by the cylinder [tex]z = 4  x^2[/tex], on the sides by the cylinder [tex]x^2 + y^2 = 4[/tex], and below by the xyplane. Here is what I have: [tex]\int_{2}^{2}\int_{\sqrt{4  x^2}}^{\sqrt{4  x^2}}\int_{0}^{4  x^2}\;dz\;dy\;dx\;=\;12\pi[/tex] Is that correct? I didn't post the many steps for integration, but the integral calulation is correct, I just need to know if I set up the integral right. Thanks again 


#2
Apr1305, 09:06 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,564

Yes, that's correct. Of course, it would be easier to do the integration in cylindrial coordinates.



#3
Apr1305, 11:25 PM

P: 492

Thanks for the double checking! The next problem uses this same integral and assumes that [tex]f\left(x, y, z\right) = \sqrt{x\;y\;z}[/tex]. Then it says to setup the integral to find the average value of the function within that solid.
Here is what I have: [tex]\frac{1}{12\pi}\;\int_{2}^{2}\int_{\sqrt{4  x^2}}^{\sqrt{4  x^2}}\int_{0}^{4  x^2}\;\sqrt{x\;y\;z}\;dz\;dy\;dx[/tex] Does that look right? 


Register to reply 
Related Discussions  
Hieght of a volume in a cylinder on its side, with known volume.  Precalculus Mathematics Homework  2  
Double integral of volume bounded by plane and paraboloid  Calculus & Beyond Homework  5  
In a race down inclined plane why does a cube reach bottom first?  Classical Physics  5  
Volume of a Region bounded by two surfaces  Calculus  3  
Volume of region bounded by cone and parabloid  Calculus  5 