Question

The graph of the equation 2x=y^2 from A(0,0) to B(2,2) is rotated around the x axis. The surface area of the resulting solid is?

Answer 1

Here's what I have so far: http://www.wolframalpha.com/input/?i=y^2%3D2x

This needs to be put in terms of x so that, when we integrate we can use dx instead of dy, correct? Also, I had decided that surface area could be found by adding up all the circumferences of the discs from 0 to 2 times dx.

Answer 2

\[y^2=2x\] turns into \[y=\sqrt(2x)\]

Answer 3

Assuming the idea of circumferences is correct, C= 2πr and \[\int\limits_{0}^{2}2\pi(\sqrt{2x})dx\]

Answer 4

your method is not quite right
you need ds first

Answer 5

What is ds?

Answer 6

I would so it the other way:\[x=\frac{y^2}2\] and\[ds=\sqrt{1+(\frac{dx}{dy})^2}\]

Answer 7

Uhhh. I don't understand that. maybe I missed something in the lesson...

Answer 8

correction:\[ds=\sqrt{1+(\frac{dx}{dy})^2}dy\]and the formula for the surface area is then\[S=\int2\pi yds\]

Answer 9

Yep. I definitely missed something important. thanks for the help. maybe if I review my materials it will look remotely familiar.

Answer 10

ds is the arc length differential
arc length is\[L=\int ds=\int\sqrt{1+(\frac{dy}{dx})}dx\]you need to understand this first before you can understand surface area

Answer 11

(2*PI)/3 * ((5*5^0.5)-1)

Answer 12

correction\[L=\int ds=\int\sqrt{1+(\frac{dy}{dx})^2}dx\]

Answer 13

the latex seems to be crowding up ...

Answer 14

here's so you can get a feel for arc length
http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength.aspx
and here is the idea extended to surface area
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

Answer 15

@amistre I switched to the SVG math render and that has fixed that problem, as well as a few others

Answer 16

Thank you.