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Find the volume of the solid of revolution formed by rotating
about the x-axis each region bounded by the given curves.

f(x) =(1/3)x + 2, y = 0, x = 1, x = 3


Sagot :

The volume of the solid thus formed is [tex]\frac{26}{27} \pi[/tex] .

Let us think of any region that is enclosed by a function g(x) and the x-axis between x=a and x=b. If this region is rotated about the x-axis, the solid of rotation that results can be thought of as consisting of thin cylindrical discs with a width of dx and a radius of g (x). Therefore, by integrating the volume of this disc, the whole volume of the solid will be revealed.

Volume=[tex]\int^{a}_{b}\pi [g(x)]^2dx[/tex]

The given function is g(x)=(1/3)x+2 ranging from x=1 to x=3

Therefore volume of the solid formed is:

[tex]\int^{3}_{1}\pi [\frac{1}{3}x]^2dx\\=\int^{3}_{1}\pi [\frac{1}{9}x^2]dx\\=\pi \int^{3}_{1} [\frac{1}{9}x^2]dx\\\\=\pi [\frac{1}{27}x^3 ]^3_1\\= \frac{26}{27} \pi[/tex]

The volume of the solid thus formed is [tex]\frac{26}{27} \pi[/tex] .

To learn more about volume of solids around a curve:

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