Refer to the figure shown below.
The shaded area is common to the given curves.
The curves y=x and y = 1/x intersect at (1,1).
The area of the shaded region is
[tex]A = \int_{1}^{2} \, (x- \frac{1}{x})dx =[ \frac{x^{2}}{2}-ln(x)]_{1}^{2} = \frac{3}{2}-ln(2) =0. 8069[/tex]
If the centroid of the shaded area is (x₁, y₁), then
[tex]0.8069x_{1} = \int_{1}^{2} x(x- \frac{1}{x})dx =[ \frac{x^{3}}{3}-x]_{1}^{2} = \frac{7}{3}-1= 1.3333\\x_{1} =1.6524[/tex]
Also,
The curve y = 1/x intersects x=2 at y = 1/2.
Therefore
[tex]0.8069y_{1} = \int_{1/2}^{1} y(2- \frac{1}{y} )dy + \int_{1}^{2} y(2-y)dy \\ = [y^{2}-y]_{1/2}^{1} + [y^{2}- \frac{y^{3}}{3}]_{1}^{2} =0.9167 \\ y_{1} =1.1361[/tex]
Answer:
The centroid is located at (1.6524, 1.1361)
