Science:Math Exam Resources/Courses/MATH101/April 2014/Question 05/Solution 1

First, the given region is the following:

On the region of interest $0\leq x\leq {\frac {\pi }{4}}$, we have $\cos(x)\geq \sin(x)$. By the formula for the centroid, we have

${\begin{aligned}{\bar {x}}={\frac {1}{A}}\int _{0}^{\frac {\pi }{4}}x(\cos x-\sin x)dx,\quad {\text{and}}\quad {\bar {y}}={\frac {1}{2A}}\int _{0}^{\frac {\pi }{4}}\cos ^{2}x-\sin ^{2}xdx\end{aligned}}$

where $A$ is the area of the region which we will compute first:

${\begin{aligned}A=\int _{0}^{\frac {\pi }{4}}\cos x-\sin xdx=\left[\sin x+\cos x\right]_{0}^{\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {2}}}-1={\sqrt {2}}-1.\end{aligned}}$

Next we will compute the integral for the $x$ portion of the centroid,

$\int _{0}^{\frac {\pi }{4}}x(\cos x-\sin x){\textrm {d}}x=\int _{0}^{\frac {\pi }{4}}x\cos x{\textrm {d}}x-\int _{0}^{\frac {\pi }{4}}x\sin x{\textrm {d}}x.$

We integrate each term by parts. For the first integral,

$\int _{0}^{\frac {\pi }{4}}x\cos x{\textrm {d}}x=\left[x\sin x\right]_{0}^{\frac {\pi }{4}}-\int _{0}^{\frac {\pi }{4}}\sin x{\textrm {d}}x=\left[x\sin x\right]_{0}^{\frac {\pi }{4}}+\left[\cos x\right]_{0}^{\frac {\pi }{4}}={\frac {\pi }{4}}{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {2}}}-1.$

For the second integral,

$\int _{0}^{\frac {\pi }{4}}x\sin x{\textrm {d}}x=-\left[x\cos x\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}\cos x{\textrm {d}}x=-\left[x\cos x\right]_{0}^{\frac {\pi }{4}}+\left[\sin x\right]_{0}^{\frac {\pi }{4}}=-{\frac {\pi }{4}}{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {2}}}.$

subtracting the two integrals we have

$\int _{0}^{\frac {\pi }{4}}x(\cos x-\sin x){\textrm {d}}x={\frac {\pi }{2{\sqrt {2}}}}-1.$

Next we need to consider the integral for the $y$ component of the centroid,

$\int _{0}^{\frac {\pi }{4}}\cos ^{2}x-\sin ^{2}x{\textrm {d}}x=\int _{0}^{\frac {\pi }{4}}\cos 2x{\textrm {d}}x,$

where we have used the trigonometric identity $\cos ^{2}x-\sin ^{2}x=\cos 2x$. Therefore we have

$\int _{0}^{\frac {\pi }{4}}\cos 2x{\textrm {d}}x=\left[{\frac {1}{2}}-0\right]={\frac {1}{2}}.$

Thus, the centroid $({\bar {x}},{\bar {y}})$ is

${\begin{aligned}{\bar {x}}={\frac {1}{A}}\int _{0}^{\frac {\pi }{4}}x(\cos x-\sin x)dx,={\frac {1}{{\sqrt {2}}-1}}\left({\frac {\pi }{2{\sqrt {2}}}}-1\right)\end{aligned}}$

and

${\begin{aligned}{\bar {y}}={\frac {1}{2A}}\int _{0}^{\frac {\pi }{4}}\cos ^{2}x-\sin ^{2}xdx={\frac {1}{2({\sqrt {2}}-1)}}\cdot {\frac {1}{2}}={\frac {1}{4({\sqrt {2}}-1)}}.\end{aligned}}$