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.

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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.