In cylindrical coordinates, we have 
, so that
correspond to the upper and lower halves of a sphere with radius 
. In spherical coordinates, this sphere is 
.
means our region is between two cylinders with radius 1 and . In spherical coordinates, the inner cylinder has equation
This cylinder meets the sphere when
which occurs at
where 
. Then 
.
The volume element transforms to
Putting everything together, we have
Graph D, because no x-value repeats and it passes the vertical line test.
Answer:
a = 1565217.39 ft / s ^ 2
t = 0.001725 seconds
Step-by-step explanation:
The first thing is to use the same system of units therefore we will pass the 28 inches to feet, like this:
28 in * (1 ft / 12 in) = 2.33 ft
Now yes, we can continue, we have the following data:
vi = 0
vf = 2700 ft / s
the equations in this case are as follows:
vf = vi + a * t
vf = a * t
rearranging for a
a = vf / t (1)
now with the position equation we know that:
x = vi * t + (a * t ^ 2) / 2
x = (a * t ^ 2) / 2 (2)
now replacing (1) in (2), we are left with:
x = (vf / t) * (t ^ 2) / 2
knowing that x would be 2.33 ft, which is when the cannonball exits the cannon.
2.33 = 2700 * t / 2
t = 2.33 * 2/2700 = 0.001725 seconds.
and now replace in (1)
a = vf / t = 2700 / 0.001725 = 1565217.39 ft / s ^ 2
(3+11i)/(3+11i)=1
So that means you can multiply 6/(3+11i) by (3+11i)/(3+11i)
Then 6(3+11i)/1
= 18+66i
