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

- March 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 - March 18th 2008, 12:02 PMTheEmptySet
- March 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

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

We obviously integrate over x.

- March 18th 2008, 12:50 PMTheEmptySetI'm blindQuote:

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

I did not

That was just plain lazy (Worried) - March 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