Answer to Question #222242 in Calculus for Besh

Question

Find the surface area generated by revolving about x-axis the area in the second quadrant under the curve y=e^x.

Answer 1

$A=2\pi\int^0_{-\infty}e^x\sqrt{1+e^{2x}}dx=[u=e^x,du=e^xdx]=2\pi\int^1_{-\infty}\sqrt{1+u^2}du=$

$=[u=sinh(v),du=cosh(v)dv]=2\pi\int^{ln(1+\sqrt{2})}_{-\infty}cosh^2(v)dv=$

$=(\frac{v}{2}+\frac{sinh(2v)}{4})|^{v=ln(1+\sqrt{2})}_{v=-\infty}= \frac{\sqrt{2}}{2}+\frac{ln(1+\sqrt{2})}{2}$