# Thread: Areas in the Plane (Calculus)

1. ## Areas in the Plane (Calculus)

Can someone help me with these problems and explain each step. Any help is appreciated.

Find the area of the regions enclosed by the lines and curves.

x = (tany)^2 and x = -(tany)^2, -Π/4 ≤y ≤ Π/4

2. Originally Posted by turtle
Can someone help me with these problems and explain each step. Any help is appreciated.
Find the area of the regions enclosed by the lines and curves.
x = (tany)^2 and x = -(tany)^2, -Π/4 ≤y ≤ Π/4
Hello,

the enclosed area is calculated by the difference of functions:

$x_1=\tan^2(y)$ and $x_2=-\tan^2(y)$

$x_1-x_2=2\tan^2(y)$

The enclosed area is:

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{2\tan^2(y) dy} = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{2\cdot \frac{\sin^2(y)}{\cos^2(y)} dy}$ $=2\cdot \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{ \frac{1-\cos^2(y)}{\cos^2(y)} dy}=2\cdot \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}{ \left(\frac{1}{\cos^2(y)}-1 \right) dy}=$
$\left[2 \cdot \tan(y) - 2y \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=\left(2-\frac{\pi}{2}\right)-\left( -2-\frac{-\pi}{2} \right)=4-\pi$

I've attached a diagram of this area.

EB