Chapter 6 - Inverse Functions - 6.2* The Natural Logarithmic Functions - 6.2* Exercises - Page 447: 77

$V=\pi\ln 2$

Using the washer method, the volume of the solid is: $V=\int_{0}^{1}\pi(\frac{1}{\sqrt{x+1}})^{2}dx$ $V=\int_{0}^{1}\pi\frac{1}{x+1}dx$ $V=\pi\int_{0}^{1}\frac{1}{x+1}dx$ $V=\pi\int_{0}^{1}\frac{(x+1)'}{x+1}dx$ Using the fudamental theorem of calculus it follows: $V=\pi[\ln(x+1)]_{0}^{1}$ $V=\pi(\ln(1+1)-\ln(0+1))$ $V=\pi\ln 2$