Center of Mass of a Plane Region with Constant Mass Density

Question

Let R be the region below the curve y = 4−x^2 to the left of y = 3x and above the x-axis. What is the centroid of the region R?

Answer 1

Area of this region is written as A equal to ∫_{(0 to 1)}(4 − x² − 3x)dx or [4x − x³/3 − 1.5x²|_{(0 to 1)}] or 13/6.

Take the moment about the y-axis as M_y = ∫_{(0 to 1)}x(4 − x² − 3x)dx which gives [2x² − 0.25 x⁴ − x³|_{(0 to 1)}] or 0.75. Then the x-coordinate of the centroid is x_bar = M_y/A or 0.75 ÷ 13/6 or 9/26.

Write the moment about the x-axis as M_x = ∫_{(0 to 3)}y(y/3)dy + ∫_{(3 to 4)}[y√(4 − y)]dy.

For ∫_{(3 to 4)}[y√(4 − y)]dy, set 4 − y = u and y = 4 − u and -dy =du. Then ∫_{(3 to 4)}[y√(4 − y)]dy translates to

-∫_{((4 − 3) to (4 − 4))}[(4 − u)√u]du.

Then obtain [y³/9|_{(0 to 3)}] + -[(8/3)u^1.5 −(2/5)u^2.5|_{(1 to 0)}] or 3 + -(-34/15) or 79/15 as M_x.

Now take y_bar as M_x/A or 79/15 ÷ 13/6 or 158/65.

The centroid sought is then (x_bar,y_bar) or (9/26,158/65).