Question

Evaluate the double integral by first identifying it as the volume of a solid. (12 − 6y)da r , r = [0, 1] × [0, 1]

Answer 1

To evaluate the double integral, we need to identify it as the volume of a solid. The given integral is (12 - 6y)dA, where r = [0, 1] x [0, 1]. We can evaluate the integral by setting up the integral as ∫(∫ (12 - 6y)dxdy) and integrating first with respect to x and then with respect to y.

Step-by-step explanation:

To evaluate the double integral, we need to identify it as the volume of a solid. The given integral is (12 - 6y)dA, where r = [0, 1] x [0, 1]. This means that we are integrating the function (12 - 6y) over the rectangular region defined by x = [0, 1] and y = [0, 1].

We can evaluate the integral by setting up the integral as follows:

∫(∫ (12 - 6y)dA) = ∫(∫ (12 - 6y)dxdy)

Now, we can integrate first with respect to x and then with respect to y:

∫(∫ (12 - 6y)dxdy) = ∫ (12x - 6xy)dx = 12x2/2 - 6xyx = 6x2 - 3xy2

Finally, we can evaluate the integral:

∫(∫ (12 - 6y)dA) = 6x2 - 3xy2 = 6(1)2 - 3(1)(1)2 = 6 - 3 = 3

Answer 2

To evaluate the double integral (12 - 6y) over the region [0, 1] × [0, 1], treat it as finding the volume of a solid. Integrate with respect to y, then x, and find that the volume is 9 cubic units.

Step-by-step explanation:

The question asks to evaluate the double integral of the function (12 - 6y) over the rectangular region [0, 1] × [0, 1]. This can be interpreted as finding the volume of a solid where the height above each point (x, y) in the region is given by the function's value at that point. The limits of integration for both the x and y variables are from 0 to 1.

To evaluate this integral, and thus find the volume:

1. Setup the double integral ∫ ∫ (12 - 6y) dy dx with the given limits [0, 1] for both x and y.
2. First, integrate with respect to y, treating x as a constant, to get ∫ (12y - 3y^2) |_0^1 dx, which simplifies to ∫ (12 - 3) dx after evaluating from y = 0 to y = 1.
3. Integrate the result with respect to x, which is simply 9x evaluated from 0 to 1.
4. The final volume is then 9(1) - 9(0), which equals 9 cubic units.