Final answer:
To evaluate the given double integral ∬ D (3x−2y)dA, where D is the region bounded by the lines y=−2x+1, y=−2x+3, y=3x/2+2, we can use a suitable change of coordinates. Let's use the change of variables x=u-v and y=u+v. The Jacobian of this transformation is 2. Now we need to find the new limits of integration for the region D in u-v plane. We rewrite the equations of the lines in terms of u and v and find that the region D is described by -1 ≤ u ≤ 1 and 0 ≤ v ≤ 4. The new double integral becomes ∬ R (3(u-v)-2(u+v))(2)dudv, where R is the new region bounded by the lines u=-1, u=1, v=0, and v=4. The integral can now be evaluated by integrating first with respect to v and then with respect to u.
Step-by-step explanation:
Detailed Answer:
To evaluate the given double integral ∬ D (3x−2y)dA, where D is the region bounded by the lines y=−2x+1, y=−2x+3, y=3x/2+2, we can use a suitable change of coordinates. Let's use the change of variables x=u-v and y=u+v. The Jacobian of this transformation is 2. Now we need to find the new limits of integration for the region D in u-v plane. We rewrite the equations of the lines in terms of u and v and find that the region D is described by -1 ≤ u ≤ 1 and 0 ≤ v ≤ 4. The new double integral becomes ∬ R (3(u-v)-2(u+v))(2)dudv, where R is the new region bounded by the lines u=-1, u=1, v=0, and v=4. The integral can now be evaluated by integrating first with respect to v and then with respect to u.
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