Answer to Question #345707 in Calculus for Gogo

Question #345707

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



1
Expert's answer
2022-05-30T16:58:34-0400
(x2)2=4(x-2)^2=4

x1=0,x2=4x_1=0, x_2=4


A=04(4(x2)2)dx=[4x(x2)33]40A=\displaystyle\int_{0}^{4}(4-(x-2)^2)dx=[4x-\dfrac{(x-2)^3}{3}]\begin{matrix} 4 \\ 0 \end{matrix}

=1683(0+83)=323=16-\dfrac{8}{3}-(0+\dfrac{8}{3})=\dfrac{32}{3}

xˉ=1A04x(4(x2)2)dx\bar{x}=\dfrac{1}{A}\displaystyle\int_{0}^{4}x(4-(x-2)^2)dx

=33204(x3+4x2)dx=332[x44+4x33]40=\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}

=332(64+2563)=2=\dfrac{3}{32}(-64+\dfrac{256}{3})=2


yˉ=1A0412((4(x2)2))2dx\bar{y}=\dfrac{1}{A}\displaystyle\int_{0}^{4}\dfrac{1}{2}((4-(x-2)^2))^2dx

=36404(x48x3+16x2)dx=\dfrac{3}{64}\displaystyle\int_{0}^{4}(x^4-8x^3+16x^2)dx

=364[x552x4+16x33]40=\dfrac{3}{64}[\dfrac{x^5}{5}-2x^4+\dfrac{16x^3}{3}]\begin{matrix} 4 \\ 0 \end{matrix}

=364(10245512+10243)=85=\dfrac{3}{64}(\dfrac{1024}{5}-512+\dfrac{1024}{3})=\dfrac{8}{5}

(2,85)(2, \dfrac{8}{5})



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment