Question

In Exercises \(15-34,\) find the area of the regions enclosed by the lines and curves. $$x=\tan ^{2} y \quad$ and $$\quad x=-\tan ^{2} y, \quad-\pi / 4 \leq y \leq \pi / 4$$

Short answer

The area of the regions enclosed by the lines and curves is \(\pi\) units squared.

Step-by-step solution

Sketch the Curves

Plot the two curves \(x=tan^2 y\) and \(x=-tan^2 y\) on a set of coordinate axes, taking note of the given range for \(y\) which is \(-\pi/4 \leq y \leq \pi/4\). From this, one can see that the curves form a shape which is symmetric about the y-axis.

Determine the Boundaries for Integration

Because the curves are perfectly symmetrical with respect to the y-axis, one only needs to find the area enclosed on one side of the y-axis, then double the result. The boundaries are thus \(y = -\pi/4\) to \(y = \pi/4\).

Set Up and Evaluate the Integral

The area between two curves is given by the definite integral \(A = \int (R-L) dy\), where (R - L) is the expression for the horizontal distance between the right curve (R) and the left curve (L). Here, the right function R is \(x=\tan^2 y\) and the left function L is \(x=-\tan^2 y\). Thus the integrand becomes \(\tan^2(y) - (-\tan^2(y)) = 2\tan^2(y)\). The integral for the area from \(-\pi/4\) to \(\pi/4\) is \(\int_{-\pi/4}^{\pi/4} 2\tan^2 y dy\). This integral simplifies to \(2 \int_{-\pi/4}^\pi/4 \sin^2 y dy - \int_{-\pi/4}^\pi/4 dy\), since \(\tan^2y = \sin^2y/(1 - \sin^2 y)\) and \(1 - \sin^2 y = \cos^2 y\). Using the power-reduction identity, the integral further simplifies to \(\int_{-\pi/4}^\pi/4 dy - \int_{-\pi/4}^\pi/4 \cos 2y dy\). The second integral equals zero, thus the integral to find is \(\int_{-\pi/4}^\pi/4 dy\). The area becomes \(2[(\pi/4) - (-\pi/4)] = \pi\).