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AP Calculus AB : How to find volume of a region

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Example Question #1 : How To Find Volume Of A Region

Consider the region bounded by the -axis, the line x=4 , and the function f(x) , where $f(x) = \sqrt{x}$ . Find the volume of the solid generated by rotating the region described about the line y=5 .

Possible Answers:

$\frac{136\pi }{3}$

$8\pi$

$\frac{160\pi ^{2}}{3}$

$\frac{160\pi }{3}$

$\frac{136\pi ^{2}}{3}$

Correct answer:

$\frac{136\pi }{3}$

Explanation:

To find the volume, one would use the washer method:

$V = \pi\int [R^{2}(x) - r^{2}(x)]dx.$

For this problem: $R(x) = 5, r(x) = 5-\sqrt{x}$ . So, the following integral would give the volume:

$V = \pi\int_{0}^{4} [5^{2} - (5-\sqrt{x})^{2} ]dx$

$V = \pi\int_{0}^{4} [25 - (25 - 10\sqrt{x}+ x)] dx$

$V = \pi\int_{0}^{4} (10\sqrt{x} - x) dx$

$V = \pi[\frac{20}{3}x^\frac{3}{2}} - \frac{x^{2}}{2}]$ , evaluated from x=0 to x=4 .

$V = \pi[\frac{20}{3}(4)\tfrac{3}{2} -\frac{4^{2}}{2} - (\frac{20}{3}(0)\tfrac{3}{2}-\frac{0^{2}}{2})]$

$V = \pi[(\frac{160}{3} - 8) - (0-0)]$

$V = \pi(\frac{160 - 24}{3}) = \frac{136\pi }{3}$

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AP Calculus AB : How to find volume of a region

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