Dated: 30-10-2024

Examples

Example

Use the double integral¹ to find the volume of tetrahedron bounded by coordinate planes² and the plane² \(z = 4 - 4x -2y\).

\[V = \iint_R (4 - 4x - 2y) dA\]

The bounding region \(R\) is a triangle and has vertices defined by equation \(0 = 4 - 4x - 2y\) and the origin.

\[2y = 4 - 4x\]

\[y = 2 - 2x\]

\[\therefore V = \int_0^1\int_0^{2 - 2x}(4 - 4x - 2y)dydx\]

\[= \int_0^1 \left[4y - 4xy - y^2\right]_0^{2 - 2x} dx\]

\[= \int_0^1 (4 - 8x + 4x^2) dx\]

\[= \frac 4 3\]

Find the volume of the solid bounded by cylinder \(x^2 + y^2= 4\) and the planes \(z + y = 4\) and \(z = 0\).

From the equation of the circle,

\[y^2 = 4 - x^2\]

\[y = \pm \sqrt{4 - x^2}\]

This means \(-\sqrt{4 - x^2} \le y \le \sqrt{4 - x^2}\)
To find \(x\), we assume \(y = 0\) and get \(-2 \le x \le 2\)
Therefore,

\[V = \int_{-2}^2\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} (4 - y) dy dx\]

Evaluate this and we will get

\[= 16\pi\]