MATH200 April 2012
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Question 08

Let E be the region in the first octant bounded by the coordinate planes, the plane
 $\displaystyle {}x+y=1,$
and the surface
 $\displaystyle {}z=y^{2}.$
Evaluate
 $\iiint \limits _{E}z{\textrm {d}}V.$

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

Use the coordinate planes and the equations $\displaystyle y=1x$ and $\displaystyle z=y^{2}$ to find the integration limits
$\displaystyle z\in [0,y^{2}]$
$\displaystyle y\in [0,1x]$
$\displaystyle x\in [0,1]$

Hint 2

Remember to integrate first with respect to depending variable(s), and later with the independent variable(s).
Hence, the integral to calculate is
$\displaystyle \int _{0}^{1}\int _{0}^{1x}\int _{0}^{y^{2}}z\;{\text{d}}z{\text{d}}y{\text{d}}x$

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Solution

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We find that the integration boundaries are $\displaystyle z\in [0,y^{2}],\ \ y\in [0,1x],\ \ x\in [0,1]$
Hence, the integral to calculate is
$\displaystyle {\begin{aligned}\int _{0}^{1}\int _{0}^{1x}\int _{0}^{y^{2}}z\;{\text{d}}z{\text{d}}y{\text{x}}&=\int _{0}^{1}\int _{0}^{1x}\left[{\frac {z^{2}}{2}}\right]_{0}^{y^{2}}\;{\text{d}}y{\text{d}}x\\&=\int _{0}^{1}\int _{0}^{1x}{\frac {y^{4}}{2}}\;{\text{d}}y{\text{d}}x\\&=\int _{0}^{1}\left[{\frac {y^{5}}{10}}\right]_{0}^{1x}\;{\text{d}}x\\&=\int _{0}^{1}{\frac {(1x)^{5}}{10}}\;{\text{d}}x\\&=\left[{\frac {(1x)^{6}}{60}}\right]_{0}^{1}\\&={\frac {1}{60}}\end{aligned}}$

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