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. $$ \int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x $$

Short 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-by-step solution

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.

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 \(\int_{0}^{2} \int_{y^{2}}^{4} d x d y\).

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.