I am working on some calc2 online problems and I seem to be stuck on one of the problems. The question reads: "Find the centroid of the region lying between the graphs of the functions $y=3\sin(x)$ and $y=3\cos(x)$ over the interval $[0,\pi/4]$." I am really lost on how to approach this problem so if someone could explain it step by step that would be greatly appreciated. Thank you so much!
[Math] Centroid of region btw $y=3\sin(x)$ and $y=3\cos(x)$ on $[0,\pi/4]$
calculus
Related Solutions
See the picture below.
As you can see, $\cos x$ is above $\sin x$ or $\cos x>\sin x$ for $\left[0,\dfrac\pi4\right]$. Therefore $$ \begin{align} M&=\int_0^\frac\pi4(\cos x-\sin x)\ dx=\sqrt{2}-1,\\\\ M_y&=\int_0^\frac\pi4x(\cos x-\sin x)\ dx=\frac14(\sqrt{2}\pi-4),\quad\Rightarrow\quad\text{use IBP} \end{align} $$ and $$ \begin{align} M_x&=\frac12\int_0^\frac\pi4(\cos x+\sin x)(\cos x-\sin x)\ dx\\ &=\frac12\int_0^\frac\pi4(\cos^2 x-\sin^2 x)\ dx\\ &=\frac12\int_0^\frac\pi4\cos2x\ dx\\ &=\frac14. \end{align} $$ I think you can handle it from here. I hope this helps.
Best Answer
Let $(x,y)$ be the center of mass. Then, let $$A=\int^{\pi/4}_0(3\cos(x)-3\sin(x))dx$$ Then, $$x=\dfrac{1}{A}\int^{\pi/4}_0x(3\cos(x)-3\sin(x))dx\\ y=\dfrac{1}{A}\int^{\pi/4}_0\dfrac{1}{2}(9\cos^2(x)-9\sin^2(x))dx$$ So, the center of mass is located at $$\left(\dfrac{\int^{\pi/4}_0x(3\cos(x)-3\sin(x))dx}{\int^{\pi/4}_0(3\cos(x)-3\sin(x))dx},\dfrac{\int^{\pi/4}_0\dfrac{1}{2}(9\cos^2(x)-9\sin^2(x))dx}{\int^{\pi/4}_0(3\cos(x)-3\sin(x))dx}\right)$$ This is easy to evaluate if you expand the integrals.
EDIT: Let's evaluate the integrals: $$A=\int^{\pi/4}_0(3\cos(x)-3\sin(x))dx=3\left(\int^{\pi/4}_0(\cos(x))dx-\int^{\pi/4}_0(\sin(x))dx\right)=3\left(\sin(x)+\cos(x)\right)|^{\pi/4}_0=3\left(\frac{2}{\sqrt{2}}\right)-3\left(1\right)=\frac{6}{\sqrt{2}}-3$$ Then, $$\int^{\pi/4}_0(3x\cos(x)-3x\sin(x))dx=\int^{\pi/4}_0(3x\cos(x))dx+\int^{\pi/4}_0(3x\sin(x))dx$$ You can show that this is $$3(x\sin(x)+\cos(x))|^{\pi/4}_0+3(\sin(x)-x\cos(x))|^{\pi/4}_0=3\left(\frac{\pi}{4\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-3+3\left(\frac{1}{\sqrt{2}}-\frac{\pi}{4\sqrt{2}}\right)=3(\sqrt{2}+1)$$ So, $$\int^{\pi/4}_0(3x\cos(x)-3x\sin(x))dx=3(\sqrt{2}+1)$$ The second integral is $$\int^{\pi/4}_0\dfrac{1}{2}(9\cos^2(x)-9\sin^2(x))dx$$ This you can show is $$\int\dfrac{1}{2}(9\cos^2(x)-9\sin^2(x))dx=\dfrac{9}{4}\left((x+\sin(x)\cos(x))+(\sin(x)\cos(x)-x)\right)=\dfrac{9}{4}\left(2\sin(x)\cos(x))\right)=\dfrac{9}{4}\sin(2x)$$ Evaluated from ${\pi/4}$ to $0$ gives $$\int^{\pi/4}_0\dfrac{1}{2}(9\cos^2(x)-9\sin^2(x))dx=\dfrac{9}{4}\sin(2x)|^{\pi/4}_0=\dfrac{9}{4}$$ So, you have the center of mass at $$\left(\dfrac{\int^{\pi/4}_0x(3\cos(x)-3\sin(x))dx}{\int^{\pi/4}_0(3\cos(x)-3\sin(x))dx},\dfrac{\int^{\pi/4}_0\dfrac{1}{2}(9\cos^2(x)-9\sin^2(x))dx}{\int^{\pi/4}_0(3\cos(x)-3\sin(x))dx}\right)= \left(\dfrac{3(\sqrt{2}+1)}{\dfrac{6}{\sqrt{2}}-3},\dfrac{9}{\dfrac{24}{\sqrt{2}}-12}\right)\approx\left(5.828427\ldots,1.810666\ldots\right)$$