Find the volume of the solid generated when the region enclosed by
y = √(x + 3), y = √(2x + 1), x = 0 is revolved about the x-axis.
Point of Intersection:x+3=2x+1x+3=2x+1,x=3−1=2V=π∫02(x+3−(2x+1)) dx=π∫02(2−x) dx=π(2x−x22)∣02=π(2)=2π\displaystyle \textsf{Point of Intersection:}\\ \sqrt{x + 3} = \sqrt{2x + 1}\\ x + 3 = 2x + 1, x = 3 - 1 = 2\\ \begin{aligned} V &= \pi \int_0^2 (x + 3 - (2x + 1))\,\, \mathrm{d}x \\&= \pi \int_0^2 (2 - x)\,\, \mathrm{d}x \\&= \pi\left(2x - \frac{x^2}{2}\right)\biggr\vert_0^2 \\&= \pi(2) = 2\pi \end{aligned}Point of Intersection:x+3=2x+1x+3=2x+1,x=3−1=2V=π∫02(x+3−(2x+1))dx=π∫02(2−x)dx=π(2x−2x2)∣∣02=π(2)=2π
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