# Areas in the Plane (Calculus)

• Feb 4th 2007, 05:05 PM
turtle
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
• Feb 4th 2007, 10:09 PM
earboth
Quote:

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:

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

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

The enclosed area is:

$\displaystyle \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}$$\displaystyle =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}=$
$\displaystyle \left[2 \cdot \tan(y) - 2y \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=\left(2-\frac{\pi}{2}\right)-\left( -2-\frac{-\pi}{2} \right)=4-\pi$

I've attached a diagram of this area.

EB