Question

surface area of revolution y=sinpix on x=0 to x=1.

Answer 1

\[y=sinpix \]

Answer 2

f'x=(cospix)pi

Answer 3

\[2\Pi \int\limits_{0}^{1}\sin Pix(\sqrt{1+(picospix)^2}\]

Answer 4

I could be mistaken, but that last integral looks to me like what I would use for arc length. I thought for area we would integrate pi(r)^2, and in your case it looks like r is sin(pix).

Answer 5

So if I am right, then you would have to square sin(pix) before integrating it. you'd have to check, but I suspect there is probably a helpful identity you could sub in for sin^2(pix)

Answer 6

finding surface area of revolution involes arc length integral

Answer 7

Arc length in 2D is very much associated with Surface Area in 3D.
The two integrals should have obvious similarities.

Answer 8

looks like i missed the surface area part, my mistake.

Answer 9

Still doesn't look right.

Answer 10

What formula is being used?

Answer 11

\[
2\pi r
\]Takes the radius and gives your circumference. Arc length isn't the radius.

Answer 12

\[\int\limits_{a}^{b}2\pi(fx)\sqrt{1+[f'(x)]^2}\]

Answer 13

this equation has been getting me the right answers so far

Answer 14

It is correct. It was also correct up at the top. Why did you give up on it?

Answer 15

not entirely sure how to square picospix, then i have to add 1 to it, then i have to to take square root. how do i go about doing that

Answer 16

is picospix squared = pi^2cos^2(pix)?

Answer 17

yh, looks like it is, just chcked in calc

Answer 18

That is fine. Don't worry about the square root. It will not bother you. Try substituting \(u = \cos(\pi x)\)

Answer 19

let u = cospix
du = sinpixdx

\[\int\limits_{a}^{b}du \sqrt{1+\pi^2u^2}\]

Answer 20

whoops, forgot the chain rule taking the derivative

Answer 21

Missing a few parts, there. \(du = -\pi\sin(\pi x)\;dx\)

Answer 22

let u=cospix
du=-pisinpix
-du/pi=sinpix
\[-1/\pi \int\limits_{a}^{b}du \sqrt{1+\pi^2u^2}\]

Answer 23

@tkhunny how about now? :p

Answer 24

I'm definitely in over my head, but that looks a little bit now like it meets sqrt(a^2 + x^2) for a trig substitution

Answer 25

Well, it's still not pretty, but you should be able to wade through it to a complete resolution. Save a whole sheet of paper for the details. :-)

Answer 26

it still looks completely un do able to me to l0l. and yeah, i think it might be a trig subsitution, but not entirely sure how i would go about do that

Answer 27

It was trig when we started. You might want to back up and try a different substitution. If all else fails, you can try Weierstrass - annoying, but it works.

Answer 28

even if i let u=sinx, i just end up with another problem with a bunch of u's and pi's that i dont know what to do with lol. guess i'll just move to another problem for now

Answer 29

thanks anyways though

Answer 30

let u =sinpix* is what i meant

Answer 31

Yea, it's not pretty, but it can be done. Good work this far and good luck on the rest!