Application of Integrals: Real-World Scenarios (8)

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Question

A water tank has the shape of the region bounded by the parabola y=x2/4y = x^2/4 and the line y=4y = 4 (with yy in metres). The tank is filled to the top with water. Find the volume of water and the area of the cross-section.

Solution — Step by Step

The parabola y=x2/4y = x^2/4 opens upward with vertex at the origin. It meets y=4y = 4 when x2/4=4x^2/4 = 4, i.e., x=±4x = \pm 4. So the cross-section is bounded between x=−4x = -4 and x=4x = 4, below by the parabola and above by y=4y = 4.

A=∫−44(4−x24)dx=2∫04(4−x24)dxA = \int_{-4}^{4}\left(4 - \frac{x^2}{4}\right) dx = 2 \int_{0}^{4}\left(4 - \frac{x^2}{4}\right) dx

=2[4x−x312]04=2(16−6412)=2(16−163)=2⋅323=643 m2= 2 \left[4x - \frac{x^3}{12}\right]_0^4 = 2\left(16 - \frac{64}{12}\right) = 2\left(16 - \frac{16}{3}\right) = 2 \cdot \frac{32}{3} = \frac{64}{3} \text{ m}^2

If the tank has length LL along the third dimension (say 5 m), volume = area × length:

V=643×5=3203≈106.67 m3V = \frac{64}{3} \times 5 = \frac{320}{3} \approx 106.67 \text{ m}^3

If the tank is formed by rotating the region around the y-axis (a paraboloid bowl), the volume is

V=∫04πx2 dy=∫04π(4y) dy=4π[y22]04=4π⋅8=32π m3≈100.53 m3V = \int_0^4 \pi x^2 \, dy = \int_0^4 \pi (4y) \, dy = 4\pi \left[\frac{y^2}{2}\right]_0^4 = 4\pi \cdot 8 = 32\pi \text{ m}^3 \approx 100.53 \text{ m}^3

Why This Works

The area between two curves is the integral of (top curve) − (bottom curve) over the relevant x-range. For the cross-section under the parabola but capped by y=4y = 4, the upper curve is the line and the lower curve is the parabola.

Volume of revolution comes from the disc method: at each height yy, the radius is x=2yx = 2\sqrt{y} (from the parabola), so the disc area is πx2=4πy\pi x^2 = 4\pi y. Integrate that area over yy from 0 to 4 to get the total volume.

Always identify which variable is your “slice direction.” Slicing horizontally (with dydy) is best when the region is bounded by curves expressed as x=f(y)x = f(y). Slicing vertically (with dxdx) is best when curves are y=g(x)y = g(x).

Alternative Method

For the area under the parabolic cap, you can also subtract: total rectangle (8×4=328 \times 4 = 32) minus the area between the parabola and the x-axis from −4-4 to 44. Area below parabola = ∫−44(x2/4)dx=32/3\int_{-4}^{4} (x^2/4) dx = 32/3. So shaded area = 32−32/3=64/332 - 32/3 = 64/3. Same answer.

Students forget that the parabola is symmetric — they integrate from −4-4 to 44 but treat it as if from 00 to 44. Either use the symmetry (multiply by 2) or integrate over the full range, not both.

Final answer: cross-section area =64/3= 64/3 m², prism volume (with L=5L = 5 m) =320/3= 320/3 m³, paraboloid volume =32π= 32\pi m³.

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