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Find the area of the surface obtained by revolving the graph of asteroid x^2/3 + y^2/3 =1 about the x-axis on the interval [0,1]

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Need Help!!

Find the area of the surface obtained by revolving the graph of asteroid x^2/3 + y^2/3 =1 about the x-axis on the interval [0,1]

anonymous · Answer

by x^2/3 you mean $$x^2/3$$ right? and the same with the y thingo?

anonymous · Answer

$$x^{2/3} + y^2/3 =1$$ i mean that

anonymous · Answer

$$x^{2/3}  + y^{2/3}  = 1$$

anonymous · Answer

oops, didn't read that properly and started working out the volume of the shape, gimmie a sec to think of surface area.

anonymous · Answer

oh ok 
lol

anonymous · Answer

Could put the equation into a y= form like$$y=\sqrt{(1-x^{2/3})^3}$$and then use the surface area integral thingo. Correct me if im wrong with rearranging

anonymous · Answer

Im not 100% sure on the complete solving of it but I can offer some ideas

anonymous · Answer

ok i ll try to solve with surface area formula

anonymous · Answer

Could be some nasty integration there though. :S