Areas in the Plane (Calculus)

**turtle** · February 4th 2007, 05:05 PM

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

**earboth** · February 4th 2007, 10:09 PM

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

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