Science:Math Exam Resources/Courses/MATH200/December 2011/Question 05 (b)

MATH200 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q6  •  Q7  •  Q8 (a)  •  Q8 (b) i  •  Q8 (b) ii  •  Q8 (b) iii  •  

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  •  April 2012  •  December 2011  •  

Question 05 (b)
Evaluate the double integral $\displaystyle \iint _{D}y{\sqrt {x^{2}+y^{2}}}\,dA$ over the region $\displaystyle D=\{(x,y)\,|\,x^{2}+y^{2}\leq 2\,\,{\mbox{and}}\,\,0\leq y\leq x\}$.

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
Make a sketch of the region of integration and consider a change of variables.

Hint 2
Change the cartesian variables $\displaystyle (x,y)$ to polar coordinates $\displaystyle (r,\phi )$, where the boundaries of the radius are $\displaystyle r\in [0,{\sqrt {2}}]$ and of the angle are $\displaystyle \phi \in \left[0,{\frac {\pi }{4}}\right]$.

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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. We use the change of variables from cartesian coordinates to polar coordinates: $\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}r\cos(\phi )\\r\sin(\phi )\end{pmatrix}}$ The region of integration in polar coordinates is $\displaystyle D=\{(r,\phi )|\ r\in [0,{\sqrt {2}}],\ \phi \in [0,{\frac {\pi }{4}}]\}$. The function $\displaystyle y{\sqrt {x^{2}+y^{2}}}$ becomes $\displaystyle r^{2}\sin(\phi )$. The volume factor for polar coordinates is $\displaystyle r$. Hence, $\displaystyle {\begin{aligned}\iint _{D}y{\sqrt {x^{2}+y^{2}}}\;{\text{d}}A&=\int _{0}^{\sqrt {2}}\int _{0}^{\frac {\pi }{4}}r^{2}\sin(\phi )r\;{\text{d}}\phi {\text{d}}r\\=&\int _{0}^{\sqrt {2}}r^{3}[-\cos(\phi )]_{0}^{\frac {\pi }{4}}\;{\text{d}}r\\=&\left[{\frac {1}{4}}r^{4}\right]_{0}^{\sqrt {2}}(-\cos({\frac {\pi }{4}})+\cos(0))\\=&1-{\frac {1}{\sqrt {2}}}\end{aligned}}$

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