Question

Help Needed Plox!

Answer 1

\[a =6, b=3 \rightarrow y=6\sin(3x^2)\]

Answer 2

Oh come on Sid, you should know this one :P haha

Answer 3

\[\int\limits_{0}^{\sqrt{\frac{ \pi}{ 3 }}} 2\pi x \left[ 6\sin(3x^2) \right]dx\]

Answer 4

Why so much work?
They just want you to pick out the correct shell, yes?

Answer 5

Oh no, we have to solve for it too, and find the volume.

Answer 6

\[2\pi \int\limits_{0}^{\sqrt{\frac{ \pi }{ 3 }}} x(6\sin3x^2)\]

Answer 7

And then I have to use Integration by parts?

Answer 8

u-sub.
You have a square x inside, and a linear x outside.

Answer 9

Oh right

Answer 10

@zepdrix Can i take the 6 out of the integral?

Answer 11

\[\large\rm 12\pi \int\limits\limits\limits_{0}^{\sqrt{\pi/3}} x\sin(3x^2)~dx\]Sure.

Answer 12

Oh, you probably don't want to though.
The 6 is going to make your u-substitution easier.\[\large\rm 2\pi \int\limits\limits\limits\limits_{0}^{\sqrt{\pi/3}} \sin(3x^2)(6x~dx)\]

Answer 13

Oh i see. okay thanks

Answer 14

@zepdrix
\[u = 3x^2, du=6x\]
\[2\pi \int\limits_{0}^{\pi} \sin(u) du\]
\[2\pi [-\cos(u)]_{0}^{\pi}\]
\[2\pi[1-1] = 0 ???\]

Answer 15

Mixing up your signs in there,\[\large\rm -2\pi\left[\cos u\right]_{0}^{\pi}\quad=\quad -2\pi\left[\cos\pi-\cos0\right]\quad=\quad-2\pi\left[-1-1\right]\]

Answer 16

O..... hehe