Find the center of mass of a thin plate of constant density covering the given region.
-The region bounded by $y = x ^ { 4 } , x = 5$, and the $x$-axis
A) $\bar { x } = \frac { 25 } { 6 } , \bar { y } = \frac { 3125 } { 18 }$
B) $\bar { x } = \frac { 35 } { 12 } , \bar { y } = \frac { 625 } { 18 }$
C) $\bar { x } = \frac { 5 } { 6 } , \bar { y } = \frac { 625 } { 18 }$
D) $\bar { x } = \frac { 25 } { 12 } , \bar { y } = \frac { 3125 } { 36 }$

Question

Quizplus · Accepted Answer

The center of mass is $\left( \frac { M_y } { M } , \frac { M_x } { M } \right)$, where $M_y = \int_a^b \int_c^d x \rho(x, y) \ dy \ dx$, $M_x = \int_a^b \int_c^d y \rho(x, y) \ dy \ dx$, and $M = \int_a^b \int_c^d \rho(x, y) \ dy \ dx$. Here, $\rho(x, y) = 1$, so$$M_y = \int_0^5 \int_0^{x^4} x \ dy \ dx = \int_0^5 \left[ xy \right]_0^{x^4} \ dx = \int_0^5 x^5 \ dx = \frac { 5^6 } { 6 }$$$$M_x = \int_0^5 \int_0^{x^4} y \ dy \ dx = \int_0^5 \left[ \frac { y^2 } { 2 } \right]_0^{x^4} \ dx = \int_0^5 \frac { x^8 } { 2 } \ dx = \frac { 5^9 } { 18 }$$$$M = \int_0^5 \int_0^{x^4} 1 \ dy \ dx = \int_0^5 x^4 \ dx = \frac { 5^5 } { 5 } = 3125$$Therefore, the center of mass is $\left( \frac { 5^6 / 6 } { 3125 } , \frac { 5^9 / 18 } { 3125 } \right) = \left( \frac { 25 } { 6 } , \frac { 3125 } { 18 } \right)$.

Find the Center of Mass of a Thin Plate of Constant