Chapter 15 - Multiple Integrals - 15.3 Double Integrals in Polar Coordinates - 15.3 Exercises - Page 1055: 26

$V$ = $6\pi$

$V$ = $\int\int[(6-x^2-y^2)-(2x^2+2y^2)]dA$ $V$ = $\int\int(6-3r^2)rdrdθ$ $V$ = $\int_0^{2\pi}\int_0^{\sqrt 2}(6r-3r^3)drdθ$ $V$ = $\int_0^{2\pi}[3r^2-\frac{3}{4}r^4]_0^{\sqrt 2}dθ$ $V$ = $\int_0^{2\pi}[(6-3)-0]dθ$ $V$ = $\int_0^{2\pi}3dθ$ $V$ = $3[θ]_0^{2\pi}$ $V$ = $6\pi$