Problem 28 - College Algebra and Calculus: An Applied Approach

Question

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x $$

Step-by-Step Solution

Answer

The sketched region is a right triangle in the first quadrant bounded by the lines \(y = 0\), \(y = \sqrt{x}\), \(x = 0\), and \(x = 4\). When the order of integration is changed, the double integral becomes \(\int_{0}^{2} \int_{y^{2}}^{4} d x d y\). Upon computation, both integrals yield an area of 4 units squared, demonstrating that both orders of integration give the same area.

Step 1: Sketch the region \(R\)

First, understand the limits of integration. The limits \(\int_{0}^{\sqrt{x}}\) imply that y varies between 0 and \(\sqrt{x}\). Likewise, the \(\int_{0}^{4}\) part implies that x varies between 0 and 4. To sketch the region, plot the lines \(y = 0\), \(y = \sqrt{x}\), \(x = 0\), and \(x = 4\). The region \(R\) is then the area bounded by these lines in the first quadrant.

Step 2: Change the order of integration

The area of region \(R\) can be described in terms of y as well. To do this, observe the graphed region. The y-values range from 0 to 2 (the square root of 4). For any given y-value in this range, x ranges from \(y^2\) to 4. The new integral is now \(\int_{0}^{2} \int_{y^{2}}^{4} d x d y\).

Step 3: Show that both orders yield the same area

To show that both orders yield the same area, compute both integrals independently. The original integral evaluates as follows: \(\int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x = \int_{0}^{4} x/2 dx = x^{2}/4]_{0}^{4} = 4\). The integral with the reversed order of variables evaluates as follows: \(\int_{0}^{2} \int_{y^{2}}^{4} d x d y = \int_{0}^{2} (4 - y^2) dy = [4y - y^{3}/3]_{0}^{2} = 4\). So indeed, we see that the area computed via the original order of integration and the reversed order of integration yield the same result of 4 units squared.