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
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
$\displaystyle 6~cos(x) = 4~sec^2(x)$
$\displaystyle cos^3(x) = \frac{2}{3}$
$\displaystyle cos(x) = \sqrt[3]{\frac{2}{3}}$
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$