1. ## 2 Volume Problems

If anyone could help with the following problems, I would greatly appreciate it. I will explain what I have done beneath the problems in case I am going in the right direction.

Find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis.
1. y = sec x , y = square root of 2 , -Π/4 ≤ x ≤ Π/4
I set R = square root of 2 and r = sec x tan x
I wasn't able to integrate sec^2 x tan^2 x (if I was on the
right track)
2. y = x^2 + 1 , y = x + 3
I set R = (x+3) - (x^2 +1) and r = 1. I don't know how to find the
intervals of integration.

2. Originally Posted by turtle
If anyone could help with the following problems, I would greatly appreciate it. I will explain what I have done beneath the problems in case I am going in the right direction.

Find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis.
1. y = sec x , y = square root of 2 , -Π/4 ≤ x ≤ Π/4
I set R = square root of 2 and r = sec x tan x
I wasn't able to integrate sec^2 x tan^2 x (if I was on the
right track)
2. y = x^2 + 1 , y = x + 3
I set R = (x+3) - (x^2 +1) and r = 1. I don't know how to find the
intervals of integration.
What do you mean by R and r?
I assume you are using vertical washers for the integrations
Thus, dV = pi(R^2 -r^2)dX.

If that is so, then why is the r for the Problem #1 be secXtanX?
That r should be secX.
So, the dV is
dV = pi[(sqrt(2))^2 -sec^2(X)]dx
dV = pi[2 -sec^2(X)]dx

Find the boundaries of dx. They are the x-values at the intersections of the horizontal line and the secant line:
secX = sqrt(2)
X = arcsec(sqrt(2)) = pi/4 or -pi/4
Therefore, the boundaries of dx are from -pi/4 to pi/4.
So,
V = (pi)INT.(-pi/4 -->pi/4)[2 -sec^2(X)]dX
V = pi[2X -tanX]|(-pi/4 --> pi/4)
V = pi[2(pi/4) -tan(pi/4) -2(-pi/4) +tan(-pi/4)]
V = pi[pi/2 -1 +pi/2 +(-1)]
V = pi[pi -2]
V = 3.5864 cubic units --------------answer.

=========================================
For the #2.

R = x+3
r = x^2 +1

dV = pi(R^2 -r^2)dx
dV = pi[(x+3)^2 -(x^2 +1)^2]dx
dV = pi[x^2 +6x +9 -(x^4 +2x^2 +1)]dx
dV = pi[x^2 +6x +9 -x^4 -2x^2 -1]dx
dV = pi[8 +6x -x^2 -x^4]dx --------------**

Boundaries for dx,
x+3 = x^2 +1
0 = x^2 -x +1 -3
x^2 -x -2 = 0
(x-2)(x+1) = 0
x = 2 or -1
So boundaries of dx are from x = -1 to x=2

Hence,
V = (pi)INT.(-1 --> 2)[8 +6x -x^2 -x^4]dx
V = pi[8x +3x^2 -(1/3)x^3 -(1/5)x^5]|(-1 --> 2)
V = pi[16 +3(4) -(1/3)(8) -(1/5)(32) -[-8 +3 -(1/3)(-1) -(1/5)(-1)]]
V = pi[28 -8/3 -32/5 +8 -3 -1/3 -1/5]
V = 23.4pi
V = 75.513 cu. units -------------------answer.