Question

please help :( calculus, surface area of a rotated curve

Answer 1

\[y=\sqrt(2x)\] from x=0 to x=7 rotated about the x-axis

Answer 2

This is how I started it: I did \[\int\limits_{0}^{7}2\pi y \sqrt(1+(\frac{ dy }{ dx })^2) dx\]

Answer 3

then I found that dy/dx is \[\frac{ 1 }{ 3 }(2x)^{3/2}\] is that right so far?

Answer 4

Unfortunately I do not have much time, but:

\[\Large y=\sqrt{2x}
\\
\frac{dy}{dx}=\frac{1}{\sqrt{2x}} \]

Answer 5

I added one to the exponent, divided by that, and divided by 2 (reverse chain rule) How come it's wrong?

Answer 6

well, you want to differentiate it, not integrate it. dy/dx means compute the derivative, not integrate.

Answer 7

oh my gosh.... I feel really stupid... haha I'm getting all mixed up. sorry, thanks for pointing that out, I should probably be able to figure it out now *facepalm*

Answer 8

that happens, no worries, yes give it a try!

Answer 9

okay, so now i have \[\int\limits_{0}^{7} 2 \pi \sqrt(2x) \sqrt(1+\frac{ 1 }{ 2x })dx\] I'm not sure which substitution I should make...

Answer 10

I swapped in the sqrt(2x) for the y

Answer 11

oh wait, I think I should probably swap in y for the 1/2x and then do substitution, that looks a bit easier

Answer 12

but then I would have to change my interval, nevermind

Answer 13

As long as you're integrating with respect to x, I would also just multiply the sqrt(2x) term in.

Answer 14

ahh... okay I'll try that thanks

Answer 15

you're welcome, best of luck, I will be back later!

Answer 16

thanks :)

Answer 17

if anyone else is there and wants to intervene with this problem, I am now down to \[2 \pi \int\limits_{0}^{7}\sqrt (2x+1) dx\]

Answer 18

I think that's right so far

Answer 19

\[\Large \checkmark \]

Integrate with an appropriate substitution and see if your result is sufficient.

Answer 20

yay, it worked! thanks :D