Problem 37 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Sketch the region of integration for the given integral and set up an equivalent integral with the order of integration reversed.\(\int_{0}^{2} \int_{0}^{4-x^{2}} f(x, y) d y d x\)

Step-by-Step Solution

Answer

The integral with the reversed order of integration is \(\int_{0}^{4} \int_{0}^{\sqrt{4-y}} f(x, y) \, dx \, dy\).

Step 1 - Identify the Bounds for Initial Integral

The integral given is \(\int_{0}^{2} \int_{0}^{4-x^{2}} f(x, y) \, dy \, dx\). The bounds for x are \(0 \leq x \leq 2\) and the bounds for y, for a fixed x, are \(0 \leq y \leq 4 - x^{2}\).

Step 2 - Sketch the Region of Integration

Draw the region based on the bounds: \(0 \leq x \leq 2\) and \(0 \leq y \leq 4 - x^{2}\). This represents the area under the parabola y = 4 - x², from x = 0 to x = 2.

Step 3 - Determine New Bounds for Reversed Order of Integration

For reversing the order of integration, solve \(y = 4 - x²\) for x, which yields \(x = \sqrt{4 - y}\) and \(x = -\sqrt{4 - y}\). In the region given, \(x\) varies between \(0\) and \sqrt{4 - y}\. Therefore, \(0 \leq y \leq 4\).

Step 4 - Set Up the Integral with Reversed Order

With the new bounds derived: y ranges from 0 to 4 and for a fix y, \(x\) goes from 0 to \sqrt{4 - y}. The integral with reversed order of integration is: \(\int_{0}^{4} \int_{0}^{\sqrt{4-y}} f(x, y) \, dx \, dy\).

Key Concepts

Order of Integration

The primary concept to understand here is the order of integration in a double integral. When working with double integrals, there are two different orders in which you can integrate: one where you integrate with respect to y first, and one where you integrate with respect to x first. In our original problem, the integral is set up as \(\textbackslash int_{0}^{2} \textbackslash int_{0}^{4-x^{2}} f(x, y) dy \, dx \), indicating that we first integrate with respect to y, then x. By reversing the order, the integral becomes \(\textbackslash int_{0}^{4} \textbackslash int_{0}^{\textbackslash sqrt{4-y}} f(x, y) dx \, dy \). Reversing the order might make the integration process simpler or more convenient depending on the function you're working with.

Region of Integration

The region of integration is the area in the xy-plane over which the integral is computed. To visualize this, we sketch the region. In the original integral, the region is bounded by \(0 \textbackslash leq x \textbackslash leq 2\) and \(0 \textbackslash leq y \textbackslash leq 4 - x^{2}\). This describes an area under the parabola y = 4 - x², starting from x = 0 to x = 2. Sketching this parabolic region helps us understand the 'footprint' of the double integral. When we reverse the order, the region isn't altered geometrically but is described differently: \(0 \textbackslash leq y \textbackslash leq 4\) and \(0 \textbackslash leq x \textbackslash leq \textbackslash sqrt{4 - y}\).

Bounds of Integration

The bounds of integration are the limits that define the region over which we're integrating. In the given integral \(\textbackslash int_{0}^{2} \textbackslash int_{0}^{4-x^{2}} f(x, y) dy \, dx \), the outer integral has x ranging from 0 to 2, while the inner integral has y ranging from 0 to 4 - x². To reverse the order of integration, we need new bounds corresponding to integrating with respect to x first. We solve for x in terms of y from the equation of the parabola y = 4 - x², getting \(x = \textbackslash sqrt{4 - y}\). The new bounds are y from 0 to 4, and for a fixed y, x ranges from 0 to \textbackslash sqrt{4 - y}. This transforms our integral to \(\textbackslash int_{0}^{4} \textbackslash int_{0}^{\textbackslash sqrt{4-y}} f(x, y) dx \, dy \).