Chapter 15 - Multiple Integrals - 15.2 Exercises - Page 1012: 29

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$R=[0,2] \times [0,\frac{\pi}{4}]$ $V=\int\int_{R} xsec^{2}ydA$ $V= \int^{\frac{\pi}{4}}_{0} [\frac{x^{2}}{2}sec^{2}y]^{2}_{0} dy$ $=\int^{\frac{\pi}{4}}_{0} 2sec^{2} ydy$ $=[2tany]^{\frac{\pi}{4}}_{0}$ (since $\frac{d}{dy}tany=sec^{2}y$) $=2[1-0]$ $=2$