Water Volumes in a Curved Container | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Water depth is measured from the bottom

A bowl can be modelled by rotating a curved region about a vertical axis. To find the water volume, the lower limit is the bottom's y-coordinate; the upper limit is the water surface's y-coordinate.

This is a vertical cross-section through the rotation axis. Depth 3 cm raises the surface from y = −8 to y = −5.

Worked example

A bowl is formed from x² + y² = 100, between y = −8 and y = 0, rotated about the y-axis. One coordinate unit is 1 cm. Find the volume of water at a depth of 3 cm.

The surface is $y=-8+3=-5$ , not $y=3$ . A horizontal disc's radius squared is $x^2=100-y^2$ .

\begin{gathered}\int_{-8}^{-5}(100-y^2)\,dy\\=\left[100y-\frac{y^3}{3}\right]_{-8}^{-5}\\=171\\V=171\pi\text{ cm}^3\end{gathered}

The disc areas increase from $36\pi$ to $75\pi$ square centimetres over a depth of 3 cm. The water volume must therefore lie between $108\pi$ and $225\pi$ cubic centimetres; our result does.

Use the actual model boundaries

This bowl is only a portion of a sphere, with a flat bottom at $y=-8$ . A whole-sphere or hemisphere formula would include space outside the bowl. To find its full capacity, integrate to its rim at $y=0$ .

Your turnPractice

For the same bowl, find its full capacity and the water volume at a depth of 2 cm.

Show worked answer

\begin{gathered}V_{\text{full}}\\=\pi\left[100y-\frac{y^3}{3}\right]_{-8}^{0}\\=\frac{1888\pi}{3}\\V_{\text{water}}\\=\pi\left[100y-\frac{y^3}{3}\right]_{-8}^{-6}\\=\frac{304\pi}{3}\end{gathered}

Both volumes are in $\text{cm}^3$ . The smaller depth gives less water than in the worked example, as expected.