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Chapter 13: Problem 30

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. $$ \int_{0}^{4} \int_{\sqrt{x}}^{2} d y d x $$

Short Answer

Expert verified
The area of the region \(R\) is \(\frac{8}{3}\) square units. Changing the order of integration gives the same result, \(\frac{8}{3}\) square units.

Step by step solution

01

Sketch Region \(R\)

The given integral can be interpreted as follows: \(x\) ranges from 0 to 4, and for each \(x\), \(y\) ranges from \(\sqrt{x}\) to 2. This gives us the region \(R\) in the xy-plane defined by: \(0 \leq x \leq 4\) , \(\sqrt{x} \leq y \leq 2\). Sketching that region will give a figure enclosed by the line \(y=\sqrt{x}\) (for \(0 \leq x \leq 4\) ), the line \(y=2\), and the y-axis.
02

Change the Order of Integration

To change the order of integration, we need to consider ranges for \(y\) and \(x\) in terms of \(y\). The region \(R\) is now defined by: \(0 \leq y \leq 2\), \(0 \leq x \leq y^2\). Therefore, the equivalent integral with changed order is: \(\int_{0}^{2} \int_{0}^{y^2} d x d y\)
03

Show Both Orders yield Same Area

Now it's time to execute the integrals and compare the results. Calculate original double integral: \(\int_{0}^{4} \int_{\sqrt{x}}^{2} d y d x = \int_{0}^{4} (2 - \sqrt{x}) d x = [2x - \frac{2}{3}x^{3/2}]_{0}^{4} = \frac{8}{3}\). Calculate the integral with changed order : \(\int_{0}^{2} \int_{0}^{y^2} d x d y = \int_{0}^{2} y^2 d y = [\frac{1}{3} y^3]_{0}^{2} = \frac{8}{3}\). Both integrals resulted in \(\frac{8}{3}\), so both orders yield the same area.

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