Question

Find the volume of the region under the surface ( z=5 x⁵and above the triangle in the ( x y )-plane with corners ( (0,0,0),(4,0,0) \) and ( (0,2,0) . Question Help: Video D Post to forum

Answer 1

The volume of the region under the surface
`\(z=5x^5\)` and above the triangle in the (xy)-plane with corners (0,0,0), (4,0,0), and (0,2,0) is
`\((640)/(3)\)` cubic units.

Step-by-step explanation:

To find the volume of the region under the surface
`\(z=5x^5\)` and above the given triangle in the (xy)-plane, we can use a double integration approach. First, let's define the limits of integration. The triangle in the (xy)-plane is formed by the points (0,0), (4,0), and (0,2). Therefore, the limits of integration for (x) are from 0 to 4, and for (y), they are from 0 to 2.

The integral setup for the volume is given by:

`\[V = \int_(0)^(2) \int_(0)^(4) 5x^5 \,dx \,dy\]`

Evaluating this double integral gives the final answer
`\((640)/(3)\)` cubic units.

In the first paragraph, we set up the integral, defining the limits of integration based on the given triangle. In the second paragraph, we explicitly state the integral and perform the integration. Finally, in the third paragraph, we provide the simplified answer for the volume of the region. This approach ensures a clear and concise explanation of the solution.