\(V: z= \sqrt{18-x^2-y^2}, z= \sqrt{x^2+y^2}, x\geq 0\)
\(\int \int \int_{V}{xdxdydz}=\int_{\frac{-\pi}{4}
}^{0}{d\varphi}\int_{0}^{\frac{\pi
}{4}}{d\theta}\int_{0}^{\sqrt{18}}{rcos\varphi sin\theta r^2rsin\theta
dr}=\)
\(\int_{\frac{-\pi}{4}}^{0}{d\phi }\int_{0}^{\frac{\pi }{4}}{d\theta
}\int_{0}^{\sqrt{18}}{r^3}cos \varphi sin \theta sin\theta dr=\)
\(\int_{\frac{-\pi}{4}}^{0}81 cos\varphi d\varphi
\frac{1}{2}\int_{0}^{\frac{\pi }{4}}{(1-cos 2\theta )d\theta
=81}(\frac{\pi }{8}-\frac{1}{4})(sin(-\frac{\pi }{4}))=\)
\(\frac{81}{4}(\frac{\pi }{2}-1)\)
Рисунок 1.1
Ответ: \(\frac{81}{4}(\frac{\pi }{2}-1)\)
