<p>
[quote]
Aw, don't do all his work FOR him... guidance, not answers
[/quote]
</p>
<p>LOL...I did get lazy and didn't compute a and b. That would be left for the OP, haha...</p>
<p>See that? Some SERIOUS algebraic computations there :D</p>
<p>Don't mess it up now :O</p>
<p>lol...people love to get help from me...because I usually end up giving them the answer...</p>
<p>What the heck...might as well finish it.
sqrt(100-x^2)=6, x=-8, x=8
∫pi<em>[sqt(100-x^2)^2-6^2]dx, -8, 8)
=∫pi</em>(64-x^2)dx, -8, 8)</p>
<p>oops... i mean [pi(r)^2] and you could easily account for the that by doing integrating the function for the radius of the bigger circe (in a flat plane)^2(pi), from -R to the point at which the cylinder becomes completely encompassed... and do the same for the positive values of R</p>
<p>pi * 1048/3</p>
<p>Hmm... I remember a problem like this in Calc III. If you make equations out of the two, and change them to polar? or spherical? coordinates, you should be able to take a triple integral and get the volume. I could be aliitle off...</p>
<p>:)</p>
<p>Haha...nice one, HiWei. You are 13 minutes late though. Unless you are not on EST, of course.</p>
<p>um... dualityim, no time zones differ by 13 mins... it's just the clock lol</p>
<p>Ummm...suppose his time zone is one hour behind mine, then he would still have 43 minutes before April Fool's day ends. That's what I meant.</p>
<p>Heh, I actually just read the first post and replied...</p>
<p>and I'm in CST...</p>
<p>:)</p>
<p>So you weren't trying to be funny?</p>
<p>this thread is still open?</p>