Answer to Question #219405 in Calculus for Anuj

Question #219405

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


1
Expert's answer
2021-07-21T17:44:58-0400

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}\n \\sqrt{3} \\\\\n 0\n\\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}\n 2\\pi \\\\\n 0\n\\end{matrix}=0"



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