Question #185054

Find the centre of gravity of a thin sheet with density d(x, y) = y, bounded by the

curves y = 4x2 and x = 4.


1
Expert's answer
2021-05-07T09:20:45-0400
m=0404x2ydydxm=\displaystyle\int_{0}^4\displaystyle\int_{0}^{4x^2}ydydx

=04[y22]4x20dx=048x4dx=[8x55]40=\displaystyle\int_{0}^4[\dfrac{y^2}{2}]\begin{matrix} 4x^2 \\ 0 \end{matrix}dx=\displaystyle\int_{0}^48x^4dx=[\dfrac{8x^5}{5}]\begin{matrix} 4 \\ 0 \end{matrix}

=81925=\dfrac{8192}{5}

Mx=0404x2y2dydxM_x=\displaystyle\int_{0}^4\displaystyle\int_{0}^{4x^2}y^2dydx

=04[y33]4x20dx=0464x63dx=[64x721]40=\displaystyle\int_{0}^4[\dfrac{y^3}{3}]\begin{matrix} 4x^2 \\ 0 \end{matrix}dx=\displaystyle\int_{0}^4\dfrac{64x^6}{3}dx=[\dfrac{64x^7}{21}]\begin{matrix} 4 \\ 0 \end{matrix}

=104857621=\dfrac{1048576}{21}

My=0404x2xydydxM_y=\displaystyle\int_{0}^4\displaystyle\int_{0}^{4x^2}xydydx

=04[xy22]4x20dx=048x5dx=[8x66]40=\displaystyle\int_{0}^4[\dfrac{xy^2}{2}]\begin{matrix} 4x^2 \\ 0 \end{matrix}dx=\displaystyle\int_{0}^48x^5dx=[\dfrac{8x^6}{6}]\begin{matrix} 4 \\ 0 \end{matrix}

=163843=\dfrac{16384}{3}


xˉ=Mym=16384(5)3(8192)=103\bar{x}=\dfrac{M_y}{m}=\dfrac{16384(5)}{3(8192)}=\dfrac{10}{3}


yˉ=Mxm=1048576(5)21(8192)=64021\bar{y}=\dfrac{M_x}{m}=\dfrac{1048576(5)}{21(8192)}=\dfrac{640}{21}



(103,64021)(\dfrac{10}{3}, \dfrac{640}{21})



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