Find the centroid of the region bounded by the given curves. Select the correct answer.
$$y = 2 \sin 5 x , y = 0 , x = 0 , x = \frac { \pi } { 5 }$$
A) $( 0.3146,1.5708 )$
B) $( - 0.6283,1.5708 )$
C) $( 0.2513,0.6283 )$
D) $( - 0.3146,0.7854 )$
E) $( 0.6283,0.3146 )$

Question

Quizplus · Accepted Answer

We can use the formula for finding the centroid of a region:
$$ \bar{x} = \frac { 1 } { A } \int _ { a } ^ { b } \int _ { f _ { 1 } ( x ) } ^ { f _ { 2 } ( x ) } x d y d x \quad \text { and } \quad \bar { y } = \frac { 1 } { A } \int _ { a } ^ { b } \int _ { f _ { 1 } ( x ) } ^ { f _ { 2 } ( x ) } \frac { y ^ { 2 } } { 2 } d y d x$$
where A is the area of the region.
First, we need to find the limits of integration:
The curves intersect at $x = 0$ and $x = \frac {\pi }{5}$, so those are the limits of integration.
Next, we need to find the area:
$$A = \int _ { 0 } ^ { \frac { \pi } { 5 } } y(x) dx = \int _ { 0 } ^ { \frac { \pi } { 5 } } 2 \sin 5x dx = -\frac {1}{5} \cos 5x \Big| _ { 0 } ^ { \frac { \pi } { 5 } } = \frac {2}{5} ( \cos 0 - \cos \pi ) = \frac {4}{5}$$
Now we can find the centroid:
\begin{align*}
\bar{x} &= \frac { 1 } { A } \int _ { 0 } ^ { \frac { \pi } { 5 } } \int _ { 0 } ^ { 2 \sin 5x } x dy dx \
&= \frac { 1 } { A } \int _ { 0 } ^ { \frac { \pi } { 5 } } x^2 \Big| _ { 0 } ^ { 2 \sin 5x } dx \
&= \frac { 1 } { A } \int _ { 0 } ^ { \frac { \pi } { 5 } } 4 \sin^2 5x dx\
&= \frac { 1 } { A } \int _ { 0 } ^ { \frac { \pi } { 5 } } 2 - 2\cos 10x dx\
&= \frac { 1 } { A } \left[ 2x - \frac {1}{5} \sin 10x \right] _ { 0 } ^ { \frac { \pi } { 5 } }\
&= \frac { 1 } { A } \left( \frac {2\pi } {5} - \frac {2\sqrt {2}} {5} \right)\
&= \frac {8 - 4\sqrt {2}} {5} \approx 0.25134\
\end{align*}
\begin{align*}
\bar{y} &= \frac { 1 } { A } \int _ { 0 } ^ { \frac { \pi } { 5 } } \int _ { 0 } ^ { 2 \sin 5x } \frac { y ^ { 2 } } { 2 } dy dx\
&= \frac { 1 } { A } \int _ { 0 } ^ { \frac { \pi } { 5 } } \frac { 4 \sin ^ { 3 } 5x } { 3 } dx\
&= \frac { 1 } { A } \int _ { 0 } ^ { \frac { \pi } { 5 } } \frac { 2\sin 5x - 2\sin 15x } { 3 } dx\
&= \frac { 1 } { A } \left[ -\frac { 2 } { 15 } \cos 5x + \frac { 2 } { 45 } \cos 15x \right] _ { 0 } ^ { \frac { \pi } { 5 } }\
&= \frac { 1 } { A } \left( \frac {1}{15} - \frac {1}{45} \right)\
&= \frac {2 } {15} \approx 0.13333\
\end{align*}
Therefore, the centroid is $(\bar{x}, \bar{y}) \approx (0.25134, 0.13333)$. Thus, the answer is C.

Find the Centroid of the Region Bounded by the Given y=2sin⁡5x,y=0,x=0,x=π5y = 2 \sin 5 x , y = 0 , x = 0 , x = \frac { \pi } { 5 }y=2sin5x,y=0,x=0,x=5π​

Find the Centroid of the Region Bounded by the Given $y = 2 \sin 5 x , y = 0 , x = 0 , x = \frac { \pi } { 5 }$