Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 362: 17

$$V = 2\pi $$

$$\eqalign{ & {\text{We have }}x = \csc y,\,\,\,\,y = \frac{\pi }{4},\,\,\,y = \frac{{3\pi }}{4},\,\,\,\,x = 0 \cr & {\text{The volume of the solid can be calculated using}} \cr & V = \int_c^d {\pi \left( {{{\left[ {w\left( y \right)} \right]}^2} - {{\left[ {v\left( y \right)} \right]}^2}} \right)dy} \cr & {\text{Let }}w\left( y \right) = \csc y{\text{ and }}v\left( y \right) = 0 \cr & {\text{Thus}}{\text{,}} \cr & V = \pi \int_{\pi /4}^{3\pi /4} {\left( {{{\left[ {\csc y} \right]}^2} - {{\left[ 0 \right]}^2}} \right)dy} \cr & V = \pi \int_{\pi /4}^{3\pi /4} {{{\csc }^2}ydy} \cr & {\text{Integrating}} \cr & V = - \pi \left[ {\cot y} \right]_{\pi /4}^{3\pi /4} \cr & V = - \pi \left[ {\cot \left( {\frac{{3\pi }}{4}} \right) - \cot \left( {\frac{\pi }{4}} \right)} \right] \cr & V = - \pi \left[ { - 1 - 1} \right] \cr & V = 2\pi \cr} $$