Question

Evaluate integral integral integral yz dv where E is the region bounded by x = 2y^2 + 2z^2 - 5 and the plane x = 1

Answer 1

Answer:-

We’ll integrate in the order $dxdydz.$

$$x=1: 1=2y^2+2z^2-5$$ $$y^2+z^2=3$$

We have $Q=\{(y, z):y^2+z^2\leq3\}$ and

$$\int \int \int_E yzdV=\int \int_Q\bigg(\displaystyle\int_{2y^2+2z^2-5}^{1}yzdx\bigg)dA$$ $$=\int \int_Q(1-2y^2-2z^2+5)yzdA$$ $$=\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{\sqrt{3}}(6-2r^2)r\sin\theta r\cos\theta rdrd\theta$$ $$=\displaystyle\int_{0}^{2\pi}\sin (2\theta )\bigg[\dfrac{3r^4}{4}-\dfrac{r^6}{6}\bigg]\begin{matrix} \sqrt{3} \\ 0 \end{matrix}d\theta$$ $$=\displaystyle\int_{0}^{2\pi}\dfrac{9}{4}\sin (2\theta )d\theta$$ $$=-\dfrac{9}{8}[\cos(2\theta)]\begin{matrix} 2\pi \\ 0 \end{matrix}=0$$