Answer to Question #345707 in Calculus for Gogo

Question

Find the center of mass of the region bounded by y=(x-2)^2 and y=4.

Answer 1

$$(x-2)^2=4$$

$$x_1=0, x_2=4$$

$$A=\displaystyle\int_{0}^{4}(4-(x-2)^2)dx=[4x-\dfrac{(x-2)^3}{3}]\begin{matrix} 4 \\ 0 \end{matrix}$$

$$=16-\dfrac{8}{3}-(0+\dfrac{8}{3})=\dfrac{32}{3}$$

$$\bar{x}=\dfrac{1}{A}\displaystyle\int_{0}^{4}x(4-(x-2)^2)dx$$

$$=\dfrac{3}{32}\displaystyle\int_{0}^{4}(-x^3+4x^2)dx=\dfrac{3}{32}[-\dfrac{x^4}{4}+\dfrac{4x^3}{3}]\begin{matrix} 4 \\ 0 \end{matrix}$$

$$=\dfrac{3}{32}(-64+\dfrac{256}{3})=2$$

$$\bar{y}=\dfrac{1}{A}\displaystyle\int_{0}^{4}\dfrac{1}{2}((4-(x-2)^2))^2dx$$

$$=\dfrac{3}{64}\displaystyle\int_{0}^{4}(x^4-8x^3+16x^2)dx$$

$$=\dfrac{3}{64}[\dfrac{x^5}{5}-2x^4+\dfrac{16x^3}{3}]\begin{matrix} 4 \\ 0 \end{matrix}$$

$$=\dfrac{3}{64}(\dfrac{1024}{5}-512+\dfrac{1024}{3})=\dfrac{8}{5}$$

$(2, \dfrac{8}{5})$