<p>I think i understand the concept but i need some help. the problem is</p>
<p>Find the volume fo the solid generated the by revolving the region bounded by y= sqrt(x), y=2, and x=0 about the </p>
<p>a. x-axis
b.y-axis
c. line y=2
d. line x=4</p>
<p>thanks to anyone who can help</p>
<p>Find the volume fo the solid generated the by revolving the region bounded by y= sqrt(x), y=2, and x=0 about the</p>
<p>a. x-axis
b.y-axis
c. line y=2
d. line x=4</p>
<p>a) Ok, the easiest way is to use cylindrical shells, and do 2pi rh. Do it as r=x, h - 2-sqrt(x) and width = dx. The points are from x=0 to x=4.
2*pi * int(x(2-sqrt(x))dx) from 0 to 4.
b)About line x=4, the only thing that changes is the radius. The new radius is (4-x) because when x=0 the radius from the line x=4 is 4. And when x=4, the radius is 0 because it is on the line x=4.
For the y use pi *((2^2) - x) from 0 to 2 for b.</p>
<p>I need to go pick up chinese.</p>
<p><a href="http://putfile.com/pic.php?pic=3/8418080690.jpg&s=x10%5B/url%5D">http://putfile.com/pic.php?pic=3/8418080690.jpg&s=x10</a>
Hope it helps! I'm not sure if I'm 100% correct, cuz I just looked it up right now! lol. so sorry if it's wrong!</p>
<p>thanks guys but i havent learned shell yet, heck i havent heard of it. so im really sorry that this cant help me more. but um i can coem back tomorrow or tuesday night and say who's right. sorry again.</p>
<p>yeah, I guess I did d wrong well... I'll redo it 2morrow ^_^</p>
