Find the area of the surface generated by revolving the curve about the x-axis.

y=(1/7)³,0≤x≤7

Question

Find the area of the surface generated by revolving the curve about the x-axis.

y=(1/7)³,0≤x≤7

loser66 · Answer

hey, is it not just a cylinder ?

anonymous · Answer

In my calculator it only shows up as a line running horizontally

loser66 · Answer

\((1/7)^3= (1/343) and it is a horizontal line

anonymous · Answer

So should I use the disk method?

loser66 · Answer

limits x =0 and x= 7 are the vertical lines which limit the curve from 0 to 7 |dw:1395988695912:dw|

kainui · Answer

y is just a constant function. That basically means it's just the radius of your cylinder with a height of 7.

loser66 · Answer

hehehe, I am wrong there, have to apply the formula with the height is 7 and the wide is (1/343)^2 * pi

anonymous · Answer

A. [7([442^3/2]-1)pi]/51
B. ([442^3]/2+1)pi
C. [7([64^3/2]+1)pi]/51
D. [7([442^3/2]-1)pi]/27
E. [7([442^3/2]+1)pi]/27

Those are my answer options

anonymous · Answer

B. [(442^[3/2])+1]pi

loser66 · Answer

sorry, I don't calculate

anonymous · Answer

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