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Anettt [7]
3 years ago
9

Find the centroid of the region that is bounded below by the​ x-axis and above by the ellipse left parenthesis StartFraction x s

quared Over 4 EndFraction right parenthesis plus left parenthesis StartFraction y squared Over 9 EndFraction right parenthesis equals 1 x2 4 y2 9

Mathematics
1 answer:
Mashcka [7]3 years ago
7 0

Answer with explanation:

The equation of the ellipse is ,whose centroid we have to find is

      \rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1

The curve cuts the x axis at (2,0) and (-2,0) and y axis at (0,3) and (0,-3).

We have to find centroid of the Ellipse on the right of y axis.

Center of gravity will lie on x axis.

       \bar{x}=\frac{\int \int {x} \, dx dy}{\int \int dxdy}\\\\\bar{y}=\frac{\int \int {y} \, dx dy}{\int \int dxdy}\\\\ \bar{x}=\frac{\int\limits^2_0 {x} \, dx \int\limits^3_{-3} {1} \, dy}{\int\limits^2_0 {1} \, dx \int\limits^3_{-3} {1} \, dy}\\\\\bar{y}=0\\\\\bar{x}=\frac{(\frac{x^2}{2})\left \{ {{x=2} \atop {x=0}} \right \times (y)\left \{ {{y=3} \atop {y=-3}} \right.}{(x) \left \{ {{x=2} \atop {x=0}} \right \times (y)\left \{ {{y=3} \atop {y=-3}}}

    \bar{x}=\frac{\frac{2^2}{2} \times [3-(-3)]}{2*3}\\\\ \bar{x}=\frac{6}{6}\\\\ \bar{x}=1\\\\ \bar{y}=0\\\\ \text{Center of gravity}=(\bar{x},\bar{y})=(1,0)

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