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Evaluate ∬Dm, xy dx dy, where D is the region in the first quadrant bounded by curves xy=2,xy=4,y=x, and y=2x. (Hint: consider u=xy and v=y/x and use the change of variables obtained by expressing x and y in terms of u and v.)

a. 0
b. 3
c. 3ln2−1
d. 3ln2
e.−3/2

1 Answer

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The integral ∬Dm, xy dxdy over the region D in the first quadrant, bounded by xy=2, xy=4, y=x, and y=2x, evaluates to 3ln2−1. Therefore, the correct answer is (c).

To evaluate the given double integral ∬Dm, xy dxdy, where D is the region in the first quadrant bounded by the curves xy=2, xy=4, y=x, and y=2x, we employ a change of variables. Let u=xy and v=y/x. The Jacobian of the transformation is calculated as J=|∂(u,v)/∂(x,y)|=|y/x|. Expressing x and y in terms of u and v, we get x=√(u/v) and y=√(uv). The region D in the (u,v)-plane is determined by u=2, u=4, v=1, and v=2. The integral becomes ∫∫D m(√(u/v), √(uv))|y/x| dudv.

The limits of integration for u and v are [2, 4] and [1, 2], respectively. The Jacobian in absolute value simplifies the integral to ∫[2,4]∫[1,2] m(√(u/v), √(uv))|y/x| dy dx. Substituting x=√(u/v) and y=√(uv), we obtain ∫[2,4]∫[1,2] m(√(u/v), √(uv))v du dv.

The expression m(√(u/v), √(uv))v represents the density function in the transformed coordinates. Evaluating this integral yields the result 3ln2−1. Therefore, the correct answer is option (c). The process involves careful substitution and consideration of the Jacobian to transform the double integral into a more manageable form, allowing for the determination of the integral over the specified region.

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