Question

In Exercises \(43-50\), sketch the region \(R\) whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area. $$ \int_{-2}^{2} \int_{0}^{4-y^{2}} d x d y $$

Short answer

The area under the curve yielded by both approaches of integration is \(16/3\).

Step-by-step solution

Visualize the Region R

The student first needs to understand and visualize the region \(R\) represented by the given integral. This is a region in the xy-plane bounded by \(y = -2\), \(y = 2\), and \(x = 4 - y^2\) (which is a parabola opening to the left with its vertex at (4,0))

Evaluate Integral in Given Order

Next, the student needs to evaluate the integral in the order provided. This involves integrating with respect to \(x\) first, then with respect to \(y\). \[ \int_{-2}^{2} \int_{0}^{4-y^{2}} d x d y = \int_{-2}^{2} \left[ x \right]_{0}^{4-y^{2}} dy = \int_{-2}^{2} (4-y^{2}) dy = \left[ 4y - y^3/3 \right]_{-2}^{2} = 16/3 \]

Switch Order of Integration and Evaluate

The student then needs to switch the order of integration, change the associated integration limits accordingly, and evaluate the integral again. For this, we need to describe \(R\) in terms of \(x\) and \(y\): it is bounded by \(x=0\), \(x=4\), \(y=\sqrt{4-x}\) and \(y=-\sqrt{4-x}\). So the integral becomes: \[ \int_{0}^{4} \int_{-\sqrt{4-x}}^{\sqrt{4-x}} d y d x = \int_{0}^{4} \left[ y \right]_{-\sqrt{4-x}}^{\sqrt{4-x}} dx = \int_{0}^{4} 2\sqrt{4-x} dx = \left[ 2 \left( 4x - x^{3/2} \right) / 3 \right]_{0}^{4} = 16/3 \]

Compare Results

Finally, it's important to note that the area obtained in both cases is the same, which validates both approaches.