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January 30, 2015

January 30, 2015

Posted by **Anonymous** on Friday, December 21, 2012 at 1:35pm.

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.

- Calculus Help!! -
**Steve**, Friday, December 21, 2012 at 3:04pmthe 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

= 6π-16

- Calculus Help!! -
**Anonymous**, Friday, December 21, 2012 at 3:09pmOk, that makes sense. How would I find the answers to b and c though? I'm not even sure where to start...

- Calculus Help!! -
**Steve**, Friday, December 21, 2012 at 3:17pmyou 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.

- Calculus Help!! -
**Anonymous**, Friday, December 21, 2012 at 3:21pmP 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.

- Calculus Help!! -
**Steve**, Friday, December 21, 2012 at 3:45pmthe 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)

- Calculus Help!! -
**Anonymous**, Sunday, December 15, 2013 at 10:57pmNewsflash: 2(x-4)+ pi doesn't cross the x-axis before 0 so you can't use zero for the integral

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