Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

y=6 cos(x) , y=(2 sec(x))^2 , x= -pi/4 , x=pi/4

any help would be greatly appreciated. thanks

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- Mar 20th 2008, 06:30 AMwaite3Find Area of the region enclosed by the given curves
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

y=6 cos(x) , y=(2 sec(x))^2 , x= -pi/4 , x=pi/4

any help would be greatly appreciated. thanks - Mar 20th 2008, 06:52 AMsuperben
Are you sure it's y=(2 sec(x))^2? is 2 part of the square thing? or is is separated like this: y=2 sec^2 x

- Mar 20th 2008, 06:59 AMwaite3
its y=(2 sec(x))^2 so (2sec(x)) is all squared.

- Mar 20th 2008, 07:11 AMTKHunny
1) Find the intersections 6cos(x) = (2sec(x))^2. It may be important to note that this solution is inside the given Domain, unlike version 2 offered by superben. In any case, resist the temptation to use decimals. The exact answer is fine.

$\displaystyle x = acos\left(\frac{\sqrt[3]{18}}{3}\right)$

2) Observe symmetries so that you do not need to care about the other point of intersection for negative x.

3) The decision is an easy one. You will need two Domains of integration if you do it with respect to y. 'x' is the wiser choice. Carefully decide which is greater over the Domain of integration.

$\displaystyle 2*\int_{0}^{acos\left(\frac{\sqrt[3]{18}}{3}\right)}6\cos(x)-(2*\sec(x))^{2}\;dx$

4) Antiderivatives of both pieces are relatively simple. Your only challenge is one of trigonometry. You'll need a right triangle to reconcile the acos argument, but it can be done exactly. - Mar 20th 2008, 07:11 AMsuperben
by graphing it, you would know that y=6 cos(x) is below y=(2 sec(x))^2..

that means it is the integral of (2 sec(x))^2 - 6 cos(x) from -pi/4 to pi/4..

solving it would give you:**8 - 6*sqrt(2)**

solution is attached. (MSWord doc)