Question

Find the volume of the solid of revolution generated by revolving about the x-axis the region under the curve y=e^x + e^-x from x=0 to x=1

Answer 1

\[\int\limits_{0}^{1} t0p - (bottom) dx \]

Answer 2

$$y=e^x+e^{-x}\\A(x)=\pi(y(x))^2=\pi(e^x+e^{-x})^2=\pi(e^{2x}+e^{-2x}+2)\\V=\int_0^1A(x)\ dx=\pi\int_0^1(e^{2x}+e^{-2x}+2)\ dx=\pi\left[\int e^{2x}\ dx+\int e^{-2x}\ dx+2x\right]\Bigg|_0^1\\\ \ \ =\frac12e^{2}-\frac12e^{-2}\right+1$$