Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to \(x\)or\(y\). Draw a typical approximating rectangle and label its height and width. Then find the area of the region.

9. \(x = 1 - {y^2},\;\;\;x = {y^2} - 1\)

Short answer

The area of the region \(\frac{8}{3}\)

Step-by-step solution

Find the area of the region

\(\begin{aligned}{l}x = 1 - {y^2},x = {y^2} - 1\;\;\\\int_{ - 1}^1 {\left( {\left( {1 - {y^2}} \right) - \left( {{y^2} - 1} \right)} \right)} dy\end{aligned}\)

\( = \int_{ - 1}^1 {\left( {2 - 2{y^2}} \right)} dy\)

\( = \left( {2y - \frac{2}{3}{y^3}} \right)_{ - 1}^1\)

\( = \left( {(2 + 2) - \left( {\frac{2}{3} + \frac{2}{3}} \right)} \right)\)

\( = 4 - \frac{4}{3}\)

\( = \frac{{12}}{3} - \frac{4}{3} = \frac{8}{3}\)

Final proof

The area of the region \(\frac{8}{3}\)