Answer to Question #348182 in Calculus for Migz

Question #348182

Given the area in the first quadrant bounded by x² = 8y, the line



x=4 and the x-axis. What is the



the volume generated by revolving this



area about the y-axis?

1
Expert's answer
2022-06-07T11:56:35-0400




42=8y=>y=24^2=8y=>y=2

x2=8y,x0=>x=8yx^2=8y, x\ge0=>x=\sqrt{8y}

V=π02(42(8y)2)dyV=\pi \displaystyle\int_{0}^{2}(4^2-(\sqrt{8y})^2)dy

=π[16y4y2]20=π(3216)=\pi[16y-4y^2]\begin{matrix} 2\\ 0 \end{matrix}=\pi(32-16)

=16π(units3)=16\pi ({units}^3)


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