posted by Anonymous .
Region R is bounded by the functions f(x) = 2(x-4) + pi, g(x) = cos^-1(x/2 - 3), and the x axis.
a. What is the area of the region R?
b. Find the volume of the solid generated when region R is rotated about the x axis.
c. Find all values c for f(x) and g(x) in the closed interval p <= c <= q for which each function equals the average value in the indicated interval.
the curves intersect at (4,π)
a = ∫[0,4] 2(x-4)+π dx
+ ∫[4,8] arccos(x/2-3) dx
= x^2+(π-8)x [0,4]
+ x arccos(x/2-3) + 6 arcsin(x/2-3) - √(4-(x-6)^2)
= 4π-16 + 2π
Ok, that makes sense. How would I find the answers to b and c though? I'm not even sure where to start...
you must have some idea where to start.
see wolframalpha.com for doing integrals
v = ∫π y^2 dx for each region.
= π∫[0,4](2(x-4)+π)^2 dx + π∫[4,8](arccos(x/2-3))^2 dx
for the avg value, not sure what p and q are supposed to be.
P and Q are the x values on the graph I was given for this problem where the two functions intersect the x axis. They don't have any numerical assignments.
the average value over [0,8] is k=(6π-16)/8
There are two x values where this occurs
f(x) = k and g(x) = k
to see where these are visit wolframalpha.com and enter
plot y=2(x-4)+pi , y=arccos(x/2-3), y=(6*pi-16)/8, x=0..8
f(c) = (6π-16)/8 = 2(c-4)+π
c = 3 - π/8
g(c) = (6π-16)/8 = arccos(c/2-3)
c = 6+2cos(3π/4 - 2)
Newsflash: 2(x-4)+ pi doesn't cross the x-axis before 0 so you can't use zero for the integral