Centroid of Plane Area
Locate the centroid of the region π bounded by the given curves. Sketch the graph.
1.) π :π¦=π₯^2, π¦=2π₯^2 β3π₯
2.) π :π¦^2=4π₯, π¦=4, π¦βππ₯ππ
3.) π :π₯=2π¦βπ¦^2, π¦βππ₯ππ
4.) π :π¦=2π₯+1, π₯+π¦=7, π₯=8
1)"x=\\frac{\\iint\\limits_D xdxdy}{\\iint\\limits_D dxdy}=\\frac{\\int\\limits_0^3\\int\\limits_{2x^2-3x}^{x^2}xdydx}{\\int\\limits_0^3\\int\\limits_{2x^2-3x}^{x^2}dydx}=\\frac{\\int\\limits_0^3(x^2-2x^2+3x)xdx}{\\int\\limits_0^3(x^2-2x^2+3x)dx}=\\frac{6.75}{4.5}=1.5\\\\\ny=\\frac{\\iint\\limits_D ydxdy}{\\iint\\limits_D dxdy}=\\frac{0.5*\\int_0^3 ((x^2)^2-(2x^2-3x)^2)dx}{4.5}=\\frac{0.5*16.2}{4.5}=1.8"
2
"x=\\frac{\\iint\\limits_D xdxdy}{\\iint\\limits_D dxdy}=\\frac{\\int\\limits_0^44x\\sqrt{x}dx}{\\int\\limits_0^44\\sqrt{x}dx}=\\frac{256\/5}{64\/3}=2.4"
due to symmetry about the x-axis, y = 0
3
"x=\\frac{\\iint\\limits_D xdxdy}{\\iint\\limits_D dxdy}=\\frac{\\int_0^1 2*x*\\sqrt{1-x}dx}{\\int_0^1 2*\\sqrt{1-x}dx}=\\frac{8\/15}{4\/3}=0.4"
due to symmetry with respect to y = 1, y = 1
4
this is a triangle
"x=(1\/3)*(2+8+8)=6\\\\\ny=(1\/3)*(5 -1+17)=7"
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