Find the area of the surface generated by rotating the graph of f(x) = 2sqrt(x+3) about the x-axis from x = 0 to x = 5

Question

anonymous · Answer

@sleepyhead314

sleepyhead314 · Answer

same idea as before, go take a look at it and try to set it up :)

anonymous · Answer

110pie?

sleepyhead314 · Answer

that's what I got! :D

anonymous · Answer

i think we're wrong let me give you the formula. i think we're messing up:

anonymous · Answer

$$SA = \int\limits_{b}^{a} 2pif(x)\sqrt{1+[f'(x)]^2} dx$$

sleepyhead314 · Answer

crapppppp
darn I'm sorry :(
I was thinking volume the whole time >,<

anonymous · Answer

noooooooo omg the other questions

sleepyhead314 · Answer

yeah I'm sooo sorry :(

anonymous · Answer

what do we do now?

anonymous · Answer

could we put it into WA?

sleepyhead314 · Answer

yeah, that's what I was thinking :P

anonymous · Answer

what did you get? i'm not even sure how to type it in lol

sleepyhead314 · Answer

I'm gonna work on the previous ones first

anonymous · Answer

k thanks. omg you're better than santa <33333333

sleepyhead314 · Answer

http://www.wolframalpha.com/input/?i=2pi+times+integral+from+0+to+5+of+%282sqrt%28x%2B3%29%29sqrt%281%2B%281%2Fsqrt%283%2Bx%29%29%5E2%29%29+dx

sleepyhead314 · Answer

format was
2pi times integral from x=(lower number) to x=(higher number) of ( (equation) (sqrt( (derivative of equation)))^2 dx
derivative of the equation should be entered manually, wolfram does not register "derivative of ..." within the full equation

anonymous · Answer

hold on

anonymous · Answer

this one is good <3

sleepyhead314 · Answer

mk :) that's the format for future reference then :)