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.
I am having terrible trouble with numerous area of a region questions. How do I do this one?
The first step is to do the sketch of the region.
Then remember that $\displaystyle \int_a^bf(x)dx$ is the area under the curve y=f(x) between x=a and x=b. To get the area between the 2 curves you need to subtract the area under the lower curve from the area under the higher curve (you can tell the two apart by looking at your graph).
Have a go and post again if you have trouble
I would suggest you either made a mistake when graphing it on your calculator or when typing it in this forum. I got
When finding the area with graphs that intersect you need to calculate the areas on each side of the intersection separately. So first you need to find the x-value of the intersection, which I will call c.
Then you need to find the size of the area to the left of c and add it to the area on the right of c, swapping which curves are upper and lower in each section.
By the way, I didn't notice that you had to decide whether to integrate with x or y when I first posted. The reason I chose x was we had equations with x= constant.
If that's what you got I probably graphed it wrong in the calculator as that was copied verbatim.
Ok, so 6 cos x is the lower one, so now I have to integrate 6 sec(x)^2 - the integration of 6cosx, yes?
And I would do so with PI/4 as the upper bound and -PI/4 as the lower bound?
When you first wrote the equation in on this forum, you had $\displaystyle (6sec(x))^2$ instead of $\displaystyle 6sec^2(x)$. Using $\displaystyle 6sec^2(x)$ the graphs do intercept but the cos line is always lower.
exactly rightOk, so 6 cos x is the lower one, so now I have to integrate 6 sec(x)^2 - the integration of 6cosx, yes?
And I would do so with PI/4 as the upper bound and -PI/4 as the lower bound?