Question

Please help! Find the surface area generated by revolving y=2sqrt(x) From x=0 to x=8 about the x-axis.

Answer 1

so you need to find the arc length between 0 and 8

Answer 2

this is equal to:
\[\int\limits_{0}^{8} \sqrt{dx^2 + dy^2}\]

Answer 3

or\[\int\limits_{}^{} ds\]

Answer 4

because:
\[ds^2 = dx^2 + dy^2\]

Answer 5

true even for curves at very small distances

Answer 6

now:
\[\int\limits_{0}^{8} \sqrt{1 + (\frac{dy}{dx})^2} dx\]

Answer 7

by multiplying by:
\[dx/dx\]

Answer 8

then squaring it once brought inside the radical

Answer 9

so take the derivative of:
\[2x^{1/2}\]
to get:
\[x^{-1/2}\]

Answer 10

square:
1/x

Answer 11

add one to and square root:
\[\sqrt{x^{-1}+1}\]

Answer 12

integrate that with respect to dx

Answer 13

either I messed up or your teacher must hate you

Answer 14

haha, okay i think i got it. thank you! and i mean usually the problem he gives arnt that bad but this one about just killed me

Answer 15

pluging it in to wolfram alpha, you get:
http://www.wolframalpha.com/input/?i=integrate+sqrt%28x%5E-1+%2B1%29

Answer 16

then you have to rotate that

Answer 17

I also dont know if he wants you to find the area of the base