Problem 61 - Calculus Early Transcendentals

Question

Reverse the order of integration in the following integrals. $$\int_{0}^{1} \int_{0}^{\cos ^{-1} y} f(x, y) d x d y$$

Step-by-Step Solution

Answer

The new limits of integration when reversing the order are: $$\int_{0}^{\frac{\pi}{2}} \int_{\cos(x)}^{1} f(x, y) d y d x$$

Step 1: Understand the original limits of integration

The given integral is: $$\int_{0}^{1} \int_{0}^{\cos ^{-1} y} f(x, y) d x d y$$ These limits tell us that y varies from 0 to 1 and x varies from 0 to \(\cos^{-1}(y)\).

Step 2: Draw the region of integration

To better understand the area being integrated, let's draw the region defined by the given limits of integration. We need to consider the following bounds: 1. y = 0 2. y = 1 3. x = 0 4. x = \(\cos^{-1}(y)\) This corresponds to a region where the lower bound is x = 0 (the vertical axis), the left bound is y = 0 (the horizontal axis), the upper bound is y = 1, and the curve x = \(\cos^{-1}(y)\) delimits the right curve which gives us a triangular shape with x-values going from 0 to \(\frac{\pi}{2}\) as y goes from 0 to 1.

Step 3: Determine the new limits of integration while switching order

To reverse the order of integration, we need to describe the same region using new limits in terms of x and y. We want to integrate y first, then x. From the above description, we know that x goes from 0 to \(\frac{\pi}{2}\). In terms of y, we already know the inverse of x = \(\cos^{-1}(y)\) is y = \(\cos(x)\). Thus, along the x-axis, the lower bound of y is \(\cos(x)\), and its upper bound is 1. Now, we can rewrite the reversed integral as follows: $$\int_{0}^{\frac{\pi}{2}} \int_{\cos(x)}^{1} f(x, y) d y d x$$

Step 4: Present the reversed integral

The reversed order of integration for the given integral is: $$\int_{0}^{\frac{\pi}{2}} \int_{\cos(x)}^{1} f(x, y) d y d x$$