Find the area of the surface generaled by revolving the curve x=ey+e−y2 in the interval 0≤y≤ln⁡3 about the y-axis

Question

Cheryl King · Accepted Answer

Step 1  Given curve is,  x=ey+e−y2, 0≤y≤ln⁡3  Step 2  Required area is,   =∫0ln⁡3x dx=∫0ln⁡3(ey+e−y2)dy  Step 3  =12[ey−e−y]0ln⁡3 (∵integral ofex is ex)  =12[(eln⁡3−e−ln⁡3)−(e0−e0)]  Step 4  =12(eln⁡3−eln⁡(3)−1) (∵mlog⁡n=log⁡nm)  =12(3−13) (∵eogx=x)  =12×9−13  =12×83  =43

raefx88y · Accepted Answer

Step 1  x=ey−eyey+e−y  x(ey+e−y)=ey−e−y  xey+xe−y=ey−e−y  xe−y+e−y=ey−xey  ey−xey=xe−y+e−y  (1−x)ey=(x+1)e−y  (1−x)ey(1−x)e−y=(x+1)e−y(1−x)e−y  eye−y=x+11−x  ey×1e−y⇒x+11−x  ey×ey=x+11−x  2y=ln⁡(x+11−x)  y=12ln⁡(1+x1−x)

nick1337 · Accepted Answer

Step 1 One has; 1) x=ey−e−yey+e−y Consider the substitution of variables; u=e2y then (1) becomes; x=u−1u+1 From this one can isolate u to one side as; u=x+11−x Upon back substitution one obtains; 2) e2y=x+11−x Taking ln of  both sides of (2) obtains; 2y=ln⁡(x+11−x) or 3) y=12[ln⁡(x+1)−ln⁡(1−x)]

Find the area of the surface generaled by revolving the

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