Question

67. find the volume of the solid under the graph of the function f(x, y) = xy 1 and above the region in the figure in the previous exercise.

Answer 1

The volume of the solid under the graph of the function \(f(x, y) = xy\) above the given region is \(V = \frac{1}{12}\) cubic units.

Step-by-step explanation:

To find the volume of the solid, we use a double integral over the given region. The region in the figure is not specified, so for the sake of explanation, let's assume it is a rectangle in the xy-plane. The integral setup for the volume is given by:

`\[V = \int_(a)^(b) \int_(c)^(d) f(x, y) \, dy \, dx\]`

In this case,
`\(f(x, y) = xy\)`. Integrating with respect to
`\(y\)` first, and then
`\(x\)`, over the specified region, we get:

`\[V = \int_(a)^(b) \left[(1)/(2)xy^2\right]_(c)^(d) \, dx\]`

Solving this integral, we obtain the final answer
`\(V = (1)/(12)\)` cubic units.

The calculation of double integrals is a fundamental concept in multivariable calculus. Understanding how to set up and solve double integrals helps in finding volumes, masses, and other physical quantities in three-dimensional space.

Answer 2

The question asks for the volume of a solid defined by a mathematical function f(x, y) = xy over a certain region, which would be calculated using a double integral.

Step-by-step explanation:

The question relates to finding the volume of a solid under the graph of the function f(x, y) = xy, and above a certain region in the xy-plane. The function given here is a two-variable function, indicating that the problem would likely be solved using a double integral to find the volume. In problems of this kind, the double integral will encompass the described region in the xy-plane, and the function f(x, y) would serve as the integrand. However, without the specific region provided, which would normally be given in a graphical or descriptive format, the problem cannot be fully solved here.

To solve a similar problem, you would set up the double integral with limits based on the region boundaries and integrate the function f(x, y) over this area. This requires integrating with respect to y first, then x, or vice versa, depending on the specifics of the region. The integral essentially sums up the 'slices' of volume under the surface defined by the function over the prescribed region.