Let $\bar { y }$ denote the y-coordinate of the centroid of the region enclosed by $y = 1 - x ^ { 2 }$ and $y = 0$ Then $\bar { y }$ =

A) $\frac { 1,125 } { 8 }$
B) $\frac { 2 } { 5 }$
C) $\frac { 93 } { 35 }$
D) $\frac { 3 } { 4 }$
E) $\frac { 45 } { 28 }$

Question

Let $\bar { y }$ denote the y-coordinate of the centroid of the region enclosed by $y = 1 - x ^ { 2 }$ and $y = 0$ Then $\bar { y }$ =

A) $\frac { 1,125 } { 8 }$
B) $\frac { 2 } { 5 }$
C) $\frac { 93 } { 35 }$
D) $\frac { 3 } { 4 }$
E) $\frac { 45 } { 28 }$

Quizplus · Accepted Answer

The y-coordinate of the centroid $\bar{y}$ of a region can be found using the formula $\bar{y} = \frac{1}{A}\int y \cdot dA$, where $A$ is the area of the region. For the region bounded by $y = 1 - x^2$ and $y = 0$, the area $A$ can be found by integrating the function $1 - x^2$ from $x = -1$ to $x = 1$, since these are the points where $y = 1 - x^2$ intersects $y = 0$. The integral of $1 - x^2$ from $-1$ to $1$ gives the area $A = \int_{-1}^{1} (1 - x^2) dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = 2\left(1 - \frac{1}{3}\right) = \frac{4}{3}$.To find $\bar{y}$, we integrate $y \cdot dA = y \cdot (1 - x^2) dx$ over the same interval. This gives $\int_{-1}^{1} y(1 - x^2) dx = \int_{-1}^{1} (1 - x^2)^2 dx$. Solving this integral, we get $\left[ x - \frac{2x^3}{3} + \frac{x^5}{5} \right]_{-1}^{1} = 2\left(1 - \frac{2}{3} + \frac{1}{5}\right) = \frac{8}{15}$.Thus, $\bar{y} = \frac{1}{A} \cdot \frac{8}{15} = \frac{3}{4} \cdot \frac{8}{15} = \frac{2}{5}$.

Let yˉ\bar { y }yˉ​ Denote the Y-Coordinate of the Centroid of the Region Enclosed

Let $\bar { y }$ Denote the Y-Coordinate of the Centroid of the Region Enclosed