Science:Math Exam Resources/Courses/MATH200/December 2011/Question 06/Solution 1

Since the region of integration is a triangle with vertices $\displaystyle (1,0),\ (2,0),\ (0,2)$ the region is bounded by the function $\displaystyle y=2-x$ and $\displaystyle y=2-2x$, see sketch

We want to integrate with respect to $\displaystyle x$ first, hence, solving for x, the integration boundaries become

$\displaystyle {\begin{aligned}\textstyle {\frac {1}{2}}(2-y)&\leq x\leq 2-y\\0&\leq y\leq 2\end{aligned}}$

and the region is $\displaystyle R=\{(x,y)\in \mathbb {R} ^{2}|\ 1-{\frac {y}{2}}\leq x\leq 2-y,\ 0\leq y\leq 2\}$

To calculate $\displaystyle {\overline {y}}={\frac {\iint _{R}y\rho (x,y)\;{\text{d}}A}{\iint _{R}\rho (x,y)\;{\text{d}}A}}$ which is the $\displaystyle y$ coordinate of the center of mass, we calculate at first the integral in the numerator.

$\displaystyle {\begin{aligned}\iint _{R}y\rho (x,y)\;{\text{d}}A&=\int _{0}^{2}\int _{1-{\frac {y}{2}}}^{2-y}y^{3}\;{\text{d}}x{\text{d}}y\\&=\int _{0}^{2}y^{3}(2-y-1+{\frac {y}{2}})\;{\text{d}}y\\&=\int _{0}^{2}(y^{3}-{\frac {y^{4}}{2}})\;{\text{d}}y\\&={\frac {2^{4}}{4}}-{\frac {2^{5}}{10}}={\frac {4}{5}}\end{aligned}}$

The integral in the denominator is very similar

$\displaystyle {\begin{aligned}\iint _{R}\rho (x,y)\;{\text{d}}A=\int _{0}^{2}\int _{1-{\frac {y}{2}}}^{2-y}y^{2}\;{\text{d}}x{\text{d}}y=\int _{0}^{2}(y^{2}-{\frac {y^{3}}{2}})\;{\text{d}}y={\frac {2}{3}}\end{aligned}}$

Hence

$\displaystyle {\overline {y}}={\frac {\frac {4}{5}}{\frac {2}{3}}}={\frac {6}{5}}$