Question

Let ( mathcald = phi(mathcalr) ), where ( phi(u, v) = (u², u v) ) and ( mathcalr=[0,4] imes [1,6] ). Calculate ( iint_mathcald y da ).

Answer 1

To calculate the double integral of y over the given region, we first apply the transformation function to express the region in terms of the transformed variables. Then, we evaluate the double integral using the limits of integration and calculate the result. The value of the double integral is 3840.

Step-by-step explanation:

We are given a transformation function φ: R^2 → R^2 defined by φ(u, v) = (u², uv). The domain of the function is R^2 and the range is also R^2. We are asked to calculate the double integral of y with respect to the region defined by the range of the function.

To evaluate the double integral, we need to express the region of integration in terms of the transformed variables u and v. The given region R in R^2 is [0, 4] × [1, 6], which corresponds to the rectangle with sides parallel to the coordinate axes. Applying the transformation, the region φ(R) in the transformed plane is [(0)², (4)²] × [(0)(1), (4)(6)] = [0, 16] × [0, 24].

Now, we can evaluate the double integral of y over the region φ(R), using the limits of integration as [0, 16] for u and [0, 24] for v:

iint_mathcald y da = iint_0^16 iint_0^24 v du dv = int_0^24 int_0^16 v u du dv = (1/2)(24)(16)(16) = 3840.