Find the volume of the solid generated when the region between the graphs of h(y)=3+y2h(y)=3+y2 and w(y)=2−y2 over the interval [0,1] is revolved about the y-axis.
Find the volume of the solid generated when the region between the graphs of over the interval is revolved about the y-axis.
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multivariable-calculus
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0please any one solve this question – 2017-02-03
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2What have u tried? Also edit your question, u have written h(y) two times. – 2017-02-03
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0Find the volume of the solid generated when the region between the graphs of h(y)=3+y2(y square) and w(y) =2−y2(y square ) over the interval [0,1] is revolved about the y-axis – 2017-02-04
1 Answers
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First, note that for $0 \le y \le 1$, we have $0 < w(y) < h(y)$. That tells us that if you slice the solid of revolution along a plane of constant $y$, the cross section is an annulus with inner radius $w(y)$ and outer radius $h(y)$. We can compute the volume by integrating the volume of thin annulus slices with thickness $dy$.
$$volume = \int_0^1 \pi(h(y)^2 - w(y)^2)\,dy$$
I leave the rest of the computation to you.