Find the centroid of the region that is bounded below by the​ x-axis and above by the ellipse left parenthesis StartFraction x squared Over 4 EndFraction right parenthesis plus left parenthesis StartFraction y squared Over 9 EndFraction right parenthesis equals 1 x2 4 y2 9

Accepted Solution

Answer with explanation:The equation of the ellipse is ,whose centroid we have to find is       [tex]\rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1[/tex]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.        [tex]\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}}}[/tex]     [tex]\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)[/tex]