anonymous
  • anonymous
Find the volume formed by rotating the region enclosed by: y=x^2, x=y^2 about the line x=-5
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
tkhunny
  • tkhunny
Find the points of intersection. This will also be your integral limits with this lovely region.
anonymous
  • anonymous
how do i do that?
tkhunny
  • tkhunny
\(x^{2} = \sqrt{x}\) -- Solve!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
0 and 1 :)
tkhunny
  • tkhunny
Perfect. Now, decide disks or shells. With this well-behaved region, it really doesn't much matter. Just pick one.
anonymous
  • anonymous
disks
tkhunny
  • tkhunny
Disks it is, then. \(\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} d(Height)\) The "Height" is the x-direction, so we have: \(\pi\int\limits_{0}^{1} OuterRadius^{2} - InnerRadius^{2} dx\) Now indentify that inner and outer radii. What say you?
anonymous
  • anonymous
x^2+5
tkhunny
  • tkhunny
Well, I would prefer the more literal (-5-x^2), but it will do since we are just going to square it. Is that the inner or the outer radius?
anonymous
  • anonymous
my bad, and R right?
tkhunny
  • tkhunny
No worries. x^2 is the inner radius. Are we calling that 'r'? What is the outer radius, "R"?
anonymous
  • anonymous
isn't (-5-x^2) the outer radius? and yes
tkhunny
  • tkhunny
No, \(x^{2}\) and \(\sqrt{x}\) are a little funny on [0,1] \(\sqrt{x} \ge x^2\;for\;x\in[0,1]\)
anonymous
  • anonymous
uhm.. im not sure ;/
tkhunny
  • tkhunny
r is \((-5 - x^{2})\) R is \((-5 - \sqrt{x})\) That's all. Did you draw a picture? You will see it at a glance.
anonymous
  • anonymous
I'm still confused toward the sqrt(x) where'd that come from?
tkhunny
  • tkhunny
x = y^2 ==> y = sqrt(x) We have integration "dx", so we need the representation in terms of x.
anonymous
  • anonymous
got it ;p lol just had to see the steps..
tkhunny
  • tkhunny
Well, there you have it. Let's see what you get.
anonymous
  • anonymous
I got 109pi/30?
tkhunny
  • tkhunny
THAT is the right answer. Good work! Wading through it is FUN, isn't it?!
anonymous
  • anonymous
haha.. totally ;p
tkhunny
  • tkhunny
Once we realized what the region was, and where the square root came from, you were all over it. This is encouraging! Move on to the next one!

Looking for something else?

Not the answer you are looking for? Search for more explanations.