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

Printable View

- Mar 18th 2008, 11:29 AMwaite3Please Help: find the area of the region
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 - Mar 18th 2008, 12:02 PMTheEmptySet
$\displaystyle \int_{-\pi/4}^{\pi/4}6\cos(x)-2 \sec^{2}xdx=6\sin(x)-2\tan(x)|_{-\pi/4}^{\pi/4}$

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

$\displaystyle 6[ \frac {\sqrt{2}}{2}- \frac{- \sqrt{2}}{2}]-2[1-(-1)]=6\sqrt{2}-4 \approx 4.485$ - Mar 18th 2008, 12:15 PMtopsquark
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

$\displaystyle 6~cos(x) = 4~sec^2(x)$

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

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

Keeping in mind that cosine is an even function of x:

$\displaystyle x = \pm cos^{-1} \left ( \sqrt[3]{\frac{2}{3}} \right )$

$\displaystyle x \approx \pm 0.508285$

We obviously integrate over x.

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

$\displaystyle = (6~sin(0.508285) - 4~tan(0.508285) ) - ( 6~sin(-0.508285) - 4~tan(-0.508285) )$$\displaystyle \approx 1.38328$ - Mar 18th 2008, 12:50 PMTheEmptySetI'm blindQuote:

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

I did $\displaystyle 2\sec^2(x)$not $\displaystyle (2\sec(x))^2$

That was just plain lazy (Worried) - Mar 19th 2008, 04:40 PMwaite3Help please!
I tried the 1.3833 answer but was incorrect and the teacher said im missing the sqrt 2 so any help would be greatly appreciated. thanks