# Thread: Find the centroid of the plane region

1. ## Find the centroid of the plane region

Find the centroid of the plane region in the second quadrant that is bounded by the curve y = 3 - (3x^2)/4

Use Pappus's theorem (not integration) to find the volume of the solid swept out when the region in the problem above is revolved about the y-axis

2. Originally Posted by victorxz90044
Find the centroid of the plane region in the second quadrant that is bounded by the curve y = 3 - (3x^2)/4

Use Pappus's theorem (not integration) to find the volume of the solid swept out when the region in the problem above is revolved about the y-axis

$\displaystyle \bar{x} = \frac{\int_a^b x \cdot f(x) \, dx}{\int_a^b f(x) \, dx}$

$\displaystyle \bar{y} = \frac{\int_a^b \frac{[f(x)]^2}{2} \, dx}{\int_a^b f(x) \, dx}$

find the cm in quad I and use symmetry ...

$\displaystyle \bar{x} = \frac{\int_0^2 x\left(3-\frac{3x^2}{4}\right) \, dx}{\int_0^2 3-\frac{3x^2}{4} \, dx} = \frac{3}{4}$

so, in quad II, $\displaystyle \bar{x} = -\frac{3}{4}$

$\displaystyle \bar{y} = \frac{\int_0^2 \frac{\left(3-\frac{3x^2}{4}\right)^2}{2} \, dx}{\int_0^2 3-\frac{3x^2}{4} \, dx} = \frac{6}{5}$

using Pappas ...

$\displaystyle V = A \cdot 2\pi r$

$\displaystyle V = \int_0^2 3-\frac{3x^2}{4} \, dx \cdot \frac{3\pi}{2} = 6\pi$
$\displaystyle V = \int_0^2 3-\frac{3x^2}{4} \, dx \cdot \frac{12\pi}{5} = \frac{48\pi}{5}$