Problem 32 - College Algebra and Calculus: An Applied Approach

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

Step-by-Step Solution

Answer

The area computed by both integrals is the same and equal to \(16/3\). The region \(R\) is a semi-circle with radius 2 centered at the origin in the xy-plane.

Step 1: Identify the region and sketch it

Identify the region \(R\) from the given limits of integration: \(x\) ranges from \(0\) to \(4-y^{2}\) (which is a semi-circle), and \(y\) ranges from \(-2\) to \(2\). So, the region \(R\) can be represented as a semi-circle with radius 2 and centered at the origin in the xy-plane.

Step 2: Change the order of integration

The goal now is to express \(y\) as a function of \(x\). Observing the semi-circle, we note that given a certain \(x\), \(y\) ranges from \(-\sqrt{4-x^{2}}\) to \(\sqrt{4-x^{2}}\), whereas \(x\) ranges from \(0\) to \(2\). Thus, the double integral can be rewritten as \[ \int_{0}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} dy dx \]

Step 3: Compute the area using both orders of integration

First compute the original integral \[ \int_{-2}^{2} \int_{0}^{4-y^{2}} dx dy = \int_{-2}^{2} [x]_{0}^{4-y^{2}} dy = \int_{-2}^{2} (4-y^{2}) dy = [4y-\frac{y^{3}}{3}]_{-2}^{2} = 16/3 \] Then compute the integral with the order of integration reversed \[ \int_{0}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} dy dx = \int_{0}^{2} [y]_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} dx = \int_{0}^{2} 2\sqrt{4-x^{2}} dx = \frac{1}{2} \pi r^{2} = 16/3 \] Thus, both orders of integration yield the same area.