Question

In Exercises 9 and 10 , find the area of the region by integrating (a) with respect to \(x\) and (b) with respect to \(y\). $$ \begin{array}{l} x=4-y^{2} \\ x=y-2 \end{array} $$

Short answer

The area of the region under the curves when integrating with respect to \(x\) is \(16/3\) and when integrating with respect to \(y\) is \(25/3\).

Step-by-step solution

Find points of intersection

Firstly, set the two equations equal to each other to find the points of intersection. Solve the equation \(4 - y^{2} = y - 2\) which can be rewritten as \(y^{2} + y - 6 = 0\). This is a quadratic equation and can be solved by factoring to get \(y - 2\)(\(y + 3\) = 0. So, \(y = 2\) and \(y = -3\) are where the curves intersect.

Integrate with respect to x

For integrating with respect to \(x\), solve the equation \(x = 4 - y^{2}\) for \(y\). You will get two solutions \(y = \sqrt{4 - x}\) and \(y = -\sqrt{4 - x}\). Find the definite integral of the difference of these equations on the interval \(-2 \leq x \leq 2\). Thus \(\int_{-2}^{2} [\sqrt{4 - x} - (-\sqrt{4 - x})] dx = 16/3\) is the area under the curve when integrated with respect to \(x\).

Integrate with respect to y

For integrating with respect to \(y\), you need the lower limit as lower function and upper limit as upper function. So take \(y - 2\) as the lower limit and \(4 - y^{2}\) as the upper limit. Find the definite integral from \(-3 \leq y \leq 2\), thus \(\int_{-3}^{2} [(4 - y^{2}) - (y - 2)] dy = 25/3\) which is the area under the curve when integrated with respect to \(y\).