Find the centroid of the region bounded by the graphs of the given equations. $y=15-x^{2}, \quad y=3-x$
A) $\left(\frac{1}{2}, \frac{37}{5}\right)$
B) $\left(\frac{1}{2}, \frac{5}{2}\right)$
C) $\left(\frac{5}{2}, \frac{1}{2}\right)$
D) $\left(\frac{37}{5}, \frac{1}{2}\right)$

Question

Quizplus · Accepted Answer

The centroid of the region is given by the formula $( \overline{x}, \overline{y} ) = \left( \frac{1}{A} \int_a^b x[f(x)-g(x)] \ dx, \frac{1}{A} \int_a^b \frac{1}{2}[f(x)^2-g(x)^2] \ dx \right)$, where $A$ is the area of the region.In this case, the area of the region is given by$$A = \int_0^4 (15-x^2) - (3-x) \ dx = \int_0^4 (12-x^2) \ dx = \left[ 12x - \frac{x^3}{3} \right]_0^4 = 48$$The x-coordinate of the centroid is given by$$\overline{x} = \frac{1}{A} \int_0^4 x[(15-x^2) - (3-x)] \ dx = \frac{1}{48} \int_0^4 x(12-x^2) \ dx$$$$= \frac{1}{48} \left[ 6x^2 - \frac{x^4}{4} \right]_0^4 = \frac{1}{48} (96 - 64) = \frac{1}{2}$$The y-coordinate of the centroid is given by$$\overline{y} = \frac{1}{A} \int_0^4 \frac{1}{2}[(15-x^2)^2 - (3-x)^2] \ dx = \frac{1}{96} \int_0^4 (225-60x^2+x^4 - 9+6x-x^2) \ dx$$$$= \frac{1}{96} \int_0^4 (216-66x^2+x^4) \ dx = \frac{1}{96} \left[ 216x - 22x^3 + \frac{x^5}{5} \right]_0^4 = \frac{1}{96} (864 - 352 + 1024) = \frac{37}{5}$$Therefore, the centroid of the region is $( \overline{x}, \overline{y} ) = \left( \frac{1}{2}, \frac{37}{5} \right)$.

Find the Centroid of the Region Bounded by the Graphs y=15−x2,y=3−xy=15-x^{2}, \quad y=3-xy=15−x2,y=3−x

Find the Centroid of the Region Bounded by the Graphs $y=15-x^{2}, \quad y=3-x$