Finding the volume of water in a swimming pool

To address the question that stalled you: a circular pool isn't oval. If the problem meant for you to consider an oval pool, it would have said "an elliptical pool with (say) horizontal axis $a$ and vertical axis $b$," or something like that. You don't need polar coordinates for this, nor do you need a double integral. Single-variable calculus will do the trick just fine. For starters, orient the pool so that the circle which defines it is centered at the origin, and look down at the pool from overhead. Think of the water in it as if it was a big block of jello. Now imagine slicing that block of jello horizontally infinitely many times, so that you have a stack of jello slices, each of which is infinitely thin, and each of which has its own width and depth, and each of which is located at its own $y$ coordinate (that is to say, for each slice, the points that make up the slice all have the same $y$-coordinate). Each slice is essentially a rectangular box with volume width times depth times "thinness". In the figure below, the blue lines represent the individual jello slices, with the thick one being the slice located at $y = -3$. Using the equation of the circle (which, since it's centered at the origin, is $x^2 + y^2 = r^2$, where $r$ is the radius of the circle), you can figure out the width of each jello slice. Knowing that the depth increases linearly, you can also figure out the depth of each slice as a function of its $y$ coordinate. [Actually, it's not important that the depth increase linearly, per se, just that it changes north-south (i.e. as a function of $y$) and not east-west (i.e. as a function of $x$). If it changed east-west instead, you'd slice the jello vertically instead of horizontally. If it changed in all directions, then you'd need a double integral.]. Anyway, once you know the width and the depth, you can compute the volume of each slice: $$V_{\mbox{slice}} = \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$ where $dy$ represents the infinite "thinness" of the slice. Now it's just a question of adding them up, which is exactly the point of a definite integral. So the total volume is: $$V_{\mbox{total}} = \int_{-r}^{-r} V_{\mbox{slice}} = \int_{-r}^{r} \mbox{width}_{\mbox{slice}} \cdot \mbox{depth}_{\mbox{slice}} \cdot dy$$ I'll leave you to figure out the appropriate expressions (in terms of $y$) for the width and the depth of each slice.