The question I have is as follows: A rectangular swimming pool with vertical sides that measures 50ft by 150ft has a depth that increases in linearly from the short end (3 feet deep) to the deep end (15 feet deep). This is assuming that water weights 62.4 pounds per cubic foot. I'm trying to set up an integral to compute this and I'm just getting stuck on the z coordinate aspect of the triple integral. would it just be bounded by 3<=z<=12?
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Suppose that the $x$-axis runs parallel to the long side, with the origin at a corner where the depth is $3$. Then the depth $z$ at the point with coordinates $(x,y)$ is given by $z=3+\frac{12x}{150}$ (here $0\le x\le 150$, $0\le y\le 50$).
If we want to find the volume using a triple integral, then we are integrating $dz \,dx\,dy$, where $z$ runs from $0$ to $3+\frac{12x}{150}$, then $x$ runs from $0$ to $150$, and finally $y$ runs from $0$ to $50$. For the weight, integrate $62.4\,dz\,dx\,dy$ over the same region.
Remark: It is perhaps simpler to set up the problem using a double integral, or a one variable integration. Most simply, we can use basic geometry. A cross-section perpendicular to the $y$-axis is a trapezoid whose area is easy to calculate. Then we multiply by $50$.
André Nicolas
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yeah, unfortunately it's required that I use a triple integral. Thanks so much though! – duxrule May 15 '13 at 16:38
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The main answer sets it up as a triple integral. To write down explicitly what it says, the volume is (as an iterated integral) $\int_{y=0}{50}\int_{x=0}^{150}\int_{z=0}^{3+12x/150}dz,dx,dy$. – André Nicolas May 15 '13 at 16:42