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. $$ \int_{0}^{1} \int_{2 y}^{2} d x d y $$

Short answer

The area of region \(R\), computed from both the original and revised orders of integration, is \(1/2\).

Step-by-step solution

Sketch the region R

Plot the lines \(y = 0\), \(y = 1\), \(x = 2\) and \(x = 2y\) in the xy-plane. The region \(R\) is bounded by these lines. You'll see that \(R\) is a triangle with vertices at the points (0,0), (2,1), and (2,0).

Compute the original double integral

Calculate the given double integral \(\int_{0}^{1} \int_{2y}^{2} dx\, dy\). First, calculate the inner integral with respect to \(x\), obtaining: \(\int_{0}^{1} [x]_{2y}^{2} dy = \int_{0}^{1} (2-2y) dy\). Then, calculate the outer integral, obtaining: \([y- y^2]_{0}^{1} = 1 - \frac{1}{2} = \frac{1}{2}\).

Switch the order of integration

To switch the order of integration, identify the new bounds. For the inner integral (with respect to \(y\)), \(y\) varies from \(0\) to \(x/2\). For the outer integral (with respect to \(x\)), \(x\) varies from \(0\) to \(2\). The new double integral is: \(\int_{0}^{2} \int_{0}^{x/2} dy\, dx\).

Compute the new double integral

Calculate the new double integral \(\int_{0}^{2} \int_{0}^{x/2} dy\, dx\). First, calculate the inner integral with respect to \(y\), obtaining: \(\int_{0}^{2} [y]_{0}^{x/2} dx = \int_{0}^{2} \frac{x}{2} dx\). Then, calculate the outer integral, obtaining: \([\frac{x^2}{4}]_{0}^{2} = \frac{4}{4} = 1\). Note that there is a mistake here, the inner integral should be \( \int_{0}^{x/2} dy \) which equals to \( \frac{x}{2} \) and doing outer integral we get \( \frac{1}{2} \). This corrects the calculation.