Question

Consider the solid obtained by rotating the region bounded by the given curves about the line x=2

x=2y^2,x=2

Find the volume V of this solid
I'm having a problem with the setup

Answer 1

do you have the answer?

Answer 2

It's been a while since I did these.

Answer 3

I have some clues
The first is that since I'm rotating around a vertical line, the cross section is going to be a disk

Answer 4

I think the answer is possibly \[\int\limits_{-1}^{1} (2y^2-2)^2 dy \]

Answer 5

but no gaurantees

Answer 6

there should be a pi

Answer 7

The center of the disk is at x=2, so the radius is 2-2y^2, but I don't under stand how this was arrived at

Answer 8

I am fairly sure it doesn't matter if it is (2-2y^2) or (2y^2-2) as you sqaure the expression anyway.

Answer 9

So I have to figure out A(x) and take the integral of that but I'm lost on how to get the area of the horizontal slice

Answer 10

Not the best. You need the length of the rectangle ( which is the distance between x=2 and the graph) , how do you find the find the distance between two points, you substract them

Answer 11

Uh uh

Answer 12

on the graph, the x coordinate is x=2y^2

Answer 13

Then that distance represents the radii of the little discs?

Answer 14

How do you find the area of a disc, pi r^2

Answer 15

But this region bounded by the function and x=2 is rotated around x=2

Answer 16

then once you have the area you multiply by "dy" to get the volume.

Answer 17

so the radius in this case is pi(2y^2)^2?

Answer 18

no, the radius is (2y^2-2)

Answer 19

Then the area of the discs is = pi (2y^2-2)^2

Answer 20

Oh, right, so A(x) = pi(2y^2-2)^2?

Answer 21

yes, except I think you mean A(y)

Answer 22

Yes, A(y), sorry

Answer 23

As it is rotated about the line x=2, you intersect the graph with x=2 and these give you the bounds on y , the limits which you should integrate over.

Answer 24

-1,1? because that's where 2y^2 intersects with x=2

Answer 25

\[ Vol = \pi \int\limits_{-1}^{1} [2(y^2-1)]^2 dy\]

Answer 26

yes.

Answer 27

\[Vol = 4 \pi \int\limits_{-1}^{1} (y^2-1)^2dy\]

Answer 28

Should I not square the function before taking the integral?

Answer 29

\[Vol = 8 \pi \int\limits_{0}^{1} (y^2-1)^2 dy\]

As integrand is even.

Answer 30

lol im probably confusing you , all I am doing is simiplifying . I did sqaure it above.

Answer 31

Now I am really confused because (2y^2-2)^2 = 4y^4-4y^2+4 = 4(y^4-y^2+1)

Answer 32

no, (2y^2 -2)^2 = 4y^4 -8y^2 +4

Answer 33

Oh, duh

Answer 34

That's why I like to simplify it, minimises mistakes.

I factored out the 2 and sqaure it by itself

Answer 35

That's where the 4 comes from at the front of the integral, 2^2 .

Answer 36

But technically, I could integrate it without simplifying, just dragging pi out front

Answer 37

And then just simplify at the end when I plug in the limits of integration

Answer 38

Anyway,

\[Vol= 8 \pi \int\limits_{0}^{1} (y^4 -2y^2 +1) = 8 \pi ( \frac{1}{5} -\frac{2}{3}+1) = \frac{64 \pi}{15}\]

Answer 39

Web assign asks "The cross-sectional area of the disk is A = π(
___). When I plugged in 4y^4-8y^2+4 it gave me an error

Answer 40

It wants me to do these in pieces