Final answer:
To find the volume of the solid under the graph of f(x, y) = 41 - x² - y² above z = 5, set up a double integral with the appropriate limits in rectangular coordinates, which is based on the circular region of intersection. Calculating the integral over these limits will give the volume of the solid.
Step-by-step explanation:
To calculate the volume of the solid under the graph of the function f(x, y) = 41 - x² - y² and above the plane z = 5, first we need to find the limits of integration for the double integral in rectangular coordinates. Since we are looking at a paraboloid, we can describe the solid of interest as a region where the function is above the plane z = 5. Therefore, we are actually looking for the volume between z = 5 and z = f(x, y).
The volume can be calculated by setting up the following double integral:
\int \int (41 - x² - y² - 5) dA
where dA is the differential area element in the xy-plane. The limits of x and y need to be determined by the intersection of the paraboloid and the plane z = 5, which results in a circle. Thus, the equation for the circle is x² + y² = 36.
Using polar coordinates (r, θ), where x = rcos(θ) and y = rsin(θ), the limits for r are from 0 to 6 (the radius of the circle), and the limits for θ are from 0 to 2π. However, since the question asks for rectangular coordinates, we must express these limits in terms of x and y. The limits for x and y would be given by the equations of the circle in the rectangular coordinates:
-6 ≤ x ≤ 6 and -\sqrt{36 - x²} ≤ y ≤ \sqrt{36 - x²}
The double integral in rectangular coordinates is then:
\int_{-6}^{6}\int_{-\sqrt{36 - x²}}^{\sqrt{36 - x²}} (41 - x² - y² - 5) dy dx
Executing this integral will give us the volume of the solid in question.