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

MATH200 December 2011

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Question 06
Let R be the triangle with vertices (0,2), (1,0), and (2,0). Let R have density $\displaystyle \rho (x,y)=y^{2}$. Find ${\bar {y}}$, the y-coordinate of the center of mass of R. You do not need to find ${\bar {x}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that the y-value of the center of mass is given by $\displaystyle {\overline {y}}={\frac {\iint _{R}y\rho (x,y)\;{\text{d}}A}{\iint _{R}\rho (x,y)\;{\text{d}}A}}$

Hint 2

To find the integration boundaries of the variables $\displaystyle x,\ y$ for the region $\displaystyle R$ a sketch is very helpful On the sketch we see that the region $\displaystyle R$ is bounded by the functions $\displaystyle y=2-x$ and $\displaystyle y=2-2x$ If we do not want to split the region into two integrals, we must integrate with respect to $\displaystyle x$ first. Find the $\displaystyle y$-depending boundaries for $\displaystyle x$, such that $\displaystyle {\begin{aligned}{\underline {\ \ \ \ \ \ \ }}&\leq x\leq {\underline {\ \ \ \ \ \ \ \ }}\\0&\leq y\leq 2\end{aligned}}$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\displaystyle {\overline {y}}={\frac {\frac {4}{5}}{\frac {2}{3}}}={\frac {6}{5}}$

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