Recent content by numberonenacho

cool! Thanks a lot for all your help. ^_^

I got 2pi r^3 for final answer after dividing by two. How does that sound?

2\pi (\int \int rdrdz) 0-2R for first integral and 0-R for second one. how would I go from there? the rdr is confusing me a little.

what would the actual integrand be? rdrdz\theta r - 0-R theta - 0-2pi z - 0-R ?

theta would be 0 - 2pi? z would be the height soo let's say 0 - a what would r be?

could you help me set it up? that's the part I am really bad at. Many thanks for discussing this problem with me. =)

Well if I cut the cylinder in half diagonally, couldn't I just take the volume of the cylinder and then divide it by two? When you say cut the cylinder with two lines what do you mean? And I have no idea how to set up the integral. >.< could you get me started please?

OH I think I get it now. So bascially, its 1/2 a cylinder right? http://img77.imageshack.us/img77/8178/shapehi9.gif So side view is square, bottom is circle and other sides are triangles. Thanks! But how would I find the volume of that shape? :bugeye:

Im still not getting it.. Pictures would help. >.< So Square base and top. Sides triangle? Where would the circle go?

So when i evaluate it, I get -12pi rt6 as the answer. Am I doing it wrong? I feel like I am not doing right. If you could, could I see how someone would do a problem like this? Evaluating integrals is a little confusing for me. And then I need lots of help on the centroid part as well. Thanks~

So I've been talking with some people about this problem, but I can't seem to find the answer, or even set it up! "A region in space, when viewed from 3 different views, looks like a circle, looks like a square and looks like a triangle. Describe this object. And then use multiple integration...

Homework Statement Find the volume bounded by sphere rho = rt. 6 and the paraboloid z = x^2 + y^2 and locate the centroid of this region The attempt at a solution http://www.mathhelpforum.com/math-help/latex2/img/4deb41286077aabd94b30802f0e6a68a-1.gif So Thats the integral that I...