Question

Find the volume formed by rotating the region enclosed by:
y=x^2, x=y^2 about the line x=-5

Answer 1

Find the points of intersection. This will also be your integral limits with this lovely region.

Answer 2

how do i do that?

Answer 3

\(x^{2} = \sqrt{x}\) -- Solve!

Answer 4

0 and 1 :)

Answer 5

Perfect.

Now, decide disks or shells. With this well-behaved region, it really doesn't much matter. Just pick one.

Answer 6

disks

Answer 7

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?

Answer 8

x^2+5

Answer 9

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?

Answer 10

my bad, and R right?

Answer 11

No worries. x^2 is the inner radius. Are we calling that 'r'?

What is the outer radius, "R"?

Answer 12

isn't (-5-x^2) the outer radius? and yes

Answer 13

No, \(x^{2}\) and \(\sqrt{x}\) are a little funny on [0,1] \(\sqrt{x} \ge x^2\;for\;x\in[0,1]\)

Answer 14

uhm.. im not sure ;/

Answer 15

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.

Answer 16

I'm still confused toward the sqrt(x) where'd that come from?

Answer 17

x = y^2 ==> y = sqrt(x)

We have integration "dx", so we need the representation in terms of x.

Answer 18

got it ;p lol just had to see the steps..

Answer 19

Well, there you have it. Let's see what you get.

Answer 20

I got 109pi/30?

Answer 21

THAT is the right answer. Good work!

Wading through it is FUN, isn't it?!

Answer 22

haha.. totally ;p

Answer 23

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!