Use a double integral to find a volume
If a solid lies between z = h(x,y) and z = g(x,y), with h ≥ g on its base R, its volume is the integral over R of h − g. The height must remain nonnegative. A plane meeting the coordinate axes often produces a triangular base rather than a rectangle. Distances measured in metres give a result in cubic metres because height is multiplied by an area element.
Worked practice example
Find the volume below z = 6 − x − y and above z = 0 in the first octant. The base is bounded by x = 0, y = 0 and x + y = 6.
| Step | Working |
|---|---|
| Set the limits | 0 ≤ x ≤ 6 and 0 ≤ y ≤ 6 − x. |
| Write the volume | V = ∫ from 0 to 6 ∫ from 0 to (6 − x) (6 − x − y) dy dx. |
| Integrate in y | [(6 − x)y − y²/2] from 0 to (6 − x) = (6 − x)²/2. |
| Integrate in x | V = ½∫ from 0 to 6 (6 − x)² dx = ½ × 72 = 36 cubic units. |
Check your method
This is a tetrahedron with three perpendicular intercepts of length 6. Its geometric volume is 6 × 6 × 6 / 6 = 36, confirming the integral. Using y ≤ 6 for every x would include points outside the base.
Try it yourself
Find the volume below z = 4 − x − y in the first octant.
Show the answer and method
Use 0 ≤ x ≤ 4 and 0 ≤ y ≤ 4 − x. The inner result is (4 − x)²/2, and the final volume is 4³/6 = 32/3 cubic units.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.