Question

The region bounded by y= the Gaussian function, y=0, x=0 and x=1 and is revolved about the y-axis. Find the volume of the resulting solid. … the Gaussian function is e^-(x^2)

Answer 1

\[y=e^{-x^2}\quad,\quad x\geq0\]\[x=\sqrt{-\ln{y}}\]\[V=\pi\int\limits_0^1x^2dy=\pi\left(\int\limits_0^{1/e} 1\cdot dy+\int\limits_{1/e}^1(\sqrt{-\ln{y}})^2dy\right)=\]\[=\pi\left(\frac{1}{e}-\int\limits_{1/e}^1\ln{y}\,\,dy\right)=\pi\left(\frac{1}{e}-y(\ln{y}-1)|_{1/e}^1\right)=\]\[=\pi\left(\frac{1}{e}+y(\ln{y}-1)|_1^{1/e}\right)=\pi\left(\frac{1}{e}-\frac{2}{e}+1\right)=\pi\left(1-\frac{1}{e}\right)\]