# Area of a Region

• Jun 23rd 2008, 04:51 PM
Pikeman85
Area of a 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.
http://hosted.webwork.rochester.edu/...85de028521.png

I am having terrible trouble with numerous area of a region questions. How do I do this one?
• Jun 23rd 2008, 04:59 PM
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
• Jun 23rd 2008, 05:07 PM
Pikeman85
Well, I graphed it on my calculator and I see the point of intersection. Which one would be the lower in this context? The secant line starts higher, but ultimately is lower, so it would be the lower curve, yes?
• Jun 23rd 2008, 05:44 PM
I would suggest you either made a mistake when graphing it on your calculator or when typing it in this forum. I got
Attachment 6945

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.
• Jun 23rd 2008, 05:46 PM
Pikeman85
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?
• Jun 23rd 2008, 06:31 PM
Pikeman85
Any idea?
• Jun 24th 2008, 04:55 PM
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.

Attachment 6952

Quote:

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?
exactly right(Nod)