Question An artist has decided to finish their piece of artwork by balancing it on a fulcrum and putting it on display. The artwork has constant density and must be balanced at its centroid. The shape of the artwork was created on a computer program then casted and fabricated. The following equation was put into the computer to generate the shape: y = 3 sin(TX) + 5 bounded by x = 0, x = 2, and y = 0 Draw the Lamina in an x-y plane and put a dot where the centroid should be. Show all work and formulas you are using. The centroid is at (7,5), where IS -

y=3 \sin (\pi x)+5x=0-10x=2-10y=0-105\left(1-\frac{3}{5 \pi}, \frac{59}{20}\right)Label 9-1)curve y=3 \sin (\pi x)+5 \quad boundeal by x=0, x=2 and y c 0Nuv\begin{aligned}M & =\int_{0}^{x} \int_{0}^{y} \rho(x, y) d A \\& =\int_{0}^{2} \int_{0}^{3 \sin (\pi x+5)} \rho d y d x \\& =\rho\left[\frac{2}{0} 3 \sin (\pi x)+5 \frac{3}{8} d x\right. \\& =\rho\left[\frac{-3}{\pi} \cos (\pi x)+5 x\right] 2 \\& \left.=\rho\left[\frac{-3}{\pi}\right)+10\right] \\& \left.=10 \rho \frac{0}{\pi}\right]\end{aligned}For center of mass of the lamina.\bar{x}=\frac{1}{m} \iint x \rho(x, y) d A \begin{aligned}\bar{x} & =\frac{1}{\log }(s) \int_{0}^{2} \int_{0}^{3 \sin (\pi x)+5} x d y d x \\& =\frac{1}{\log }(3) \int_{0}^{2} x\left[\frac{+3}{2} \sin \pi(x)+5\right] d x \\& =\frac{1}{10}\left[\frac{5 x^{2}}{2}-\frac{3}{\pi} x \cos (\pi x)+\frac{3}{\pi^{2}} \sin (\pi x)\right]_{0}^{2} \\& =\frac{1}{10}\left[-\frac{6}{\pi}+10-0\right]=1-\frac{3}{5 \pi}\end{aligned}and\begin{aligned}\bar{y} & =\frac{1}{\log }(\rho) \int_{0}^{2} \int_{0}^{3 \sin (\pi x)+5} y d y d x \\& =\frac{1}{10} \int_{0}^{2} y[3 \sin (\pi x)+5] \\& =\frac{1}{10} \int_{0}^{2}\left[\frac{y^{2}}{2}\right]_{0}^{3 \sin (\pi x)+5} d x \\& =\frac{1}{10} \int_{0}^{2} \frac{[3 \sin (\pi x)+5]}{2} d x\end{aligned} \begin{array}{l}=\frac{1}{10}\left(\frac{1}{2}\right] \int_{0}^{2}\left[9 \sin ^{2}(n x)+30 \sin (\pi m)+2 r\right] d x \\\left.=\frac{1}{20} \cdot d^{2} \frac{59 x}{2}-\frac{(3 \cos (\pi x)(3 \sin (\pi x)+20)}{2 \pi}\right]_{0}^{2} \\=\frac{1}{20}[59] \\F=\frac{59}{20} \\\end{array}The center of mas gs\begin{aligned}(-\bar{x}, \overline{9}) 1-\frac{3}{5 \pi} \quad \bar{x} & =1-\frac{3}{571} \\\bar{y} & =\frac{59}{20}\end{aligned} ....... ...