4. Rewrite each of the following as an iterated double integral with the order of intergration reversed.
(a) $\int_{0}^{1} \int_{0}^{\sqrt{y}} f(x, y) d x d y$
(b) $\int_{0}^{\pi / 2} \int_{0}^{\cos x} f(x, y) d x d y$
(c) $\int_{0}^{1} \int_{c^{y}}^{c} f(x, y) d x d y+\int_{1}^{0} \int_{c=y}^{c} f(x, y) d x d y$
Answer
Box the final answers: \(\boxed{\int_{0}^{1} \int_{x^2}^{1} f(x, y) d y d x}\), \(\boxed{\int_{0}^{1} \int_{0}^{\arccos y} f(x, y) d y d x}\), and \(\boxed{\int_{0}^{1} \int_{c^x}^{1} f(x, y) d y d x + \int_{0}^{1} \int_{0}^{x} f(x, y) d y d x}\).
Step 1 ：Define the region of integration for each integral by the given inequalities.
Step 2 ：For (a), the inequalities are 0 ≤ y ≤ 1 and 0 ≤ x ≤ √y. To reverse the order of integration, express y in terms of x, which gives the inequalities 0 ≤ x ≤ 1 and x^2 ≤ y ≤ 1. Therefore, the reversed integral is \(\int_{0}^{1} \int_{x^2}^{1} f(x, y) d y d x\).
Step 3 ：For (b), the inequalities are 0 ≤ x ≤ π/2 and 0 ≤ y ≤ cos(x). To reverse the order of integration, express x in terms of y, which gives the inequalities 0 ≤ y ≤ 1 and 0 ≤ x ≤ arccos(y). Therefore, the reversed integral is \(\int_{0}^{1} \int_{0}^{\arccos y} f(x, y) d y d x\).
Step 4 ：For (c), the region of integration is defined by the inequalities 0 ≤ y ≤ 1, c^y ≤ x ≤ c and 1 ≤ y ≤ 0, y ≤ x ≤ c. To reverse the order of integration, express y in terms of x. This gives the inequalities c^x ≤ y ≤ 1 and 0 ≤ x ≤ 1 for the first integral and 0 ≤ y ≤ x and 0 ≤ x ≤ 1 for the second integral. Therefore, the reversed integral is \(\int_{0}^{1} \int_{c^x}^{1} f(x, y) d y d x + \int_{0}^{1} \int_{0}^{x} f(x, y) d y d x\).
Step 5 ：Box the final answers: \(\boxed{\int_{0}^{1} \int_{x^2}^{1} f(x, y) d y d x}\), \(\boxed{\int_{0}^{1} \int_{0}^{\arccos y} f(x, y) d y d x}\), and \(\boxed{\int_{0}^{1} \int_{c^x}^{1} f(x, y) d y d x + \int_{0}^{1} \int_{0}^{x} f(x, y) d y d x}\).
