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

I keep getting .4853 but is incorrect

Thanks

2. Originally Posted by waite3
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=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4

I keep getting .4853 but is incorrect

Thanks
$\int_{-\pi/4}^{\pi/4}6\cos(x)-2 \sec^{2}xdx=6\sin(x)-2\tan(x)|_{-\pi/4}^{\pi/4}$

$=6[\sin(\pi /4)-\sin(-\pi /4)]-2[\tan(\pi /4)-tan(-\pi /4)]$

$6[ \frac {\sqrt{2}}{2}- \frac{- \sqrt{2}}{2}]-2[1-(-1)]=6\sqrt{2}-4 \approx 4.485$

3. Originally Posted by waite3
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=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4

I keep getting .4853 but is incorrect

Thanks
See the image below for the intersection region. (This entire area fits within the required bounds, so there are no cut-offs.

The intersection points of the two curves may be found by
$6~cos(x) = 4~sec^2(x)$

$cos^3(x) = \frac{2}{3}$

$cos(x) = \sqrt[3]{\frac{2}{3}}$

Keeping in mind that cosine is an even function of x:
$x = \pm cos^{-1} \left ( \sqrt[3]{\frac{2}{3}} \right )$

$x \approx \pm 0.508285$

We obviously integrate over x.

$A = \int_{-0.508285}^{0.508285}( 6~cos(x) - 4~sec^2(x) )~dx$

$= (6~sin(0.508285) - 4~tan(0.508285) ) - ( 6~sin(-0.508285) - 4~tan(-0.508285) )$ $\approx 1.38328$

4. ## I'm blind

y=6cos(x), y=(2sec(x))^2, x=-pi/4, x=pi/4
doh!!
I did $2\sec^2(x)$not $(2\sec(x))^2$

That was just plain lazy