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. $$ \int_{0}^{1} \int_{y^{2}}^{\sqrt[3]{y}} d x d y $$

Short answer

After changing the order of integration, the integral becomes \(\int_{0}^{1/4} (\int_{\sqrt{x}}^{x^{3}} dy) dx + \int_{1/4}^{1} (\int_{\sqrt{x}}^{1} dy) dx\). When computing both integrals, they yield the same result, thereby showing that the area under the given region R is the same regardless of the order of integration.

Step-by-step solution

Visualize the Region R

First, visualize the region in the xy plane. The double integral is defined by three curves: \(x = y^2\), \(x = \sqrt[3]{y}\), and \(y = 1\), with \(y\) ranging from 0 to 1. Plot these curves to get an idea of the region.

Find Intersection Points

To change the order of integration, find the intersection point of the curves \(x = y^2\) and \(x = \sqrt[3]{y}\). To do this, we can set the two equations equal to each other and solve for \(y\) to give us \(y = 1/2\).

Change the Order of Integration

Considering the region R, we can change the order of integration, from the x innermost integral (dx dy) to the y innermost integral (dy dx). The limits on y in terms of x will be from \(y=\sqrt{x}\) to \(y=x^3\), for \(x\) between 0 and 1/4, and from \(y=\sqrt{x}\) to \(y=1\), for \(x\) between 1/4 and 1.

Compare the Results

Calculate both integrals and show that they are equivalent. This validates our changed order of integration.