Question

In Exercises \(37-42,\) sketch the region \(R\) of integration and switch the order of integration. $$ \int_{-\pi / 2}^{\pi / 2} \int_{0}^{\cos x} f(x, y) d y d x $$

Short answer

The switched order of integration for the given integral \( \int_{-\pi / 2}^{\pi / 2} \int_{0}^{\cos x} f(x, y) d y d x \) is \( \int_{0}^{1} \int_{-\arccos y}^{\arccos y} f(x, y) dx dy \).

Step-by-step solution

Understand the Given Integral

Begin by examining the given integral: \( \int_{-\pi / 2}^{\pi / 2} \int_{0}^{\cos x} f(x, y) d y d x \). Here, the limits for \( y \) range from 0 to \( \cos x \), and the limits for \( x \) range from \( -\pi/2 \) to \( \pi/2 \). This describes the region of integration \( R \).

Sketch the Region R

On the 2-dimensional Cartesian plane, sketch the curve given by \( y = \cos x \) for \( x \) in the interval \( -\pi/2 \) to \( \pi/2 \). This forms a half-wave of the cosine function, running over the top half of the unit circle. Draw a vertical line from any point of the curve down to the x-axis (the line \( y = 0 \)). This helps visualize the region \( R \) which is bounded by the x-axis below and the cosine curve above.

Switch the Order of Integration

To switch the order of integration, first need to determine the new limits for \( x \) and \( y \). The limits of \( x \) is from \( x = 0 \) to \( x = \cos y \), and \( y \) is from \( -\pi / 2 \) to \( \pi / 2 \). Therefore, the new integral is: \( \int_{0}^{1} \int_{-\arccos y}^{\arccos y} f(x, y) dx dy \). This step is crucial as it shifts the order of 'dx' and 'dy'

Write down the Final Answer

After completing the above steps, jot down the final answer. The original integral after switching the order of integration becomes: \( \int_{0}^{1} \int_{-\arccos y}^{\arccos y} f(x, y) dx dy \).