Problem 53 - Calculus Early Transcendentals

Question

Use double integrals to calculate the volume of the following regions. The tetrahedron bounded by the coordinate planes \((x=0, y=0, z=0)\) and the plane \(z=8-2 x-4 y\)

Step-by-Step Solution

Answer

Answer: The volume of the tetrahedron is \(\frac{128}{3}\) cubic units.

Step 1: Determine the limits of integration

Let's find where the plane \(z = 8 - 2x - 4y\) intersects the coordinate planes (x=0, y=0, z=0). For x=0 plane, we have: \(z = 8 - 4y\) For y=0 plane, we have: \(z = 8 - 2x\) For z=0 plane, we have: \(0 = 8 - 2x - 4y\) Now, we can find the limits of integration for x and y. For x-limits, set y=0 and z=0: \(0 = 8 - 2x\) \(x = 4\) For y-limits, set x=0 and z=0: \(0 = 8 - 4y\) \(y = 2\)

Step 2: Set up and evaluate the double integral

The volume of the tetrahedron can be calculated using the double integral: \(V = \int \int (8 - 2x - 4y) dA\) Since the tetrahedron is in the first quadrant (x>0, y>0, z>0), we can integrate in the following order: \(V = \int_0^4 \int_0^{2-\frac{1}{2}x} (8 - 2x - 4y) dy dx\) Evaluate the integral with respect to y first: \(V = \int_0^4 (-4y^2 - 2xy + 8y) \Big|_0^{2-\frac{1}{2}x} dx\) \(V = \int_0^4 (-16 + 4x + 2x^2) dx\) Evaluate the integral with respect to x: \(V = (-\frac{16}{3}x^3 + 2x^4 + \frac{2}{5}x^5) \Big|_0^4\) \(V = \frac{128}{3}\) Therefore, the volume of the tetrahedron bounded by the coordinate planes and the plane \(z = 8 - 2x - 4y\) is \(\frac{128}{3}\) cubic units.