Chapter 15 - Multiple Integrals - 15.3 Double Integrals in Polar Coordinates - 15.3 Exercises - Page 1055: 21

$V$ = $4\pi$

$V$ = $\int_0^{2\pi}\int_0^1[4-2r\cosθ-r\sinθ]rdrdθ$ $V$ = $\int_0^{2\pi}\int_0^1[4r-2r^2\cosθ-r^2\sinθ]drdθ$ $V$ = $\int_0^{2\pi}[2r^2-\frac{2}{3}r^3\cosθ-\frac{1}{3}r^3\sinθ]_0^1dθ$ $V$ = $\int_0^{2\pi}[2-\frac{2}{3}\cosθ-\frac{1}{3}\sinθ]dθ$ $V$ = $[2θ-\frac{2}{3}\sinθ+\frac{1}{3}\cosθ]_0^{2\pi}$ $V$ = $(4\pi+\frac{1}{3})-(\frac{1}{3})$ $V$ = $4\pi$