Volume of a Solid?

Question

Find the volume of the solid whose base is the region bounded by y = x^2, y =x, x = 2 and x = 3, where cross-sections perpendicular to the x-axis are squares. I got a volume of (481*pi)/30, I was wondering if some could help me and see if thats right. Thank you.

Answer 1

taking a small dx slice of the solid, perpendicular to the x axis, we have a volume with a thickness dx, and a square cross-section (square cross-section, as mentioned in the question) with sides (x^2 - x). The volume of this slice is
(x^2 - x)^2 * dx

(Square face and dx width)

Now, Integrating this between x=2 and x=3, we have

Integral( (x^2 - x)^2 dx, between x=2, x=3)
= [x^5/5 - 2* x^4/4 + x^3/3, between x=2, x=3]
= [3^5/5 - 3^4/2 + 3^3/3 - (2^5/5 - 2^4/2 + 2^3/3)]
= 481/30

Answer 2

If I remember it correctly,

taking a small dx slice of the solid, perpendicular to the x axis, we have a volume with a thickness dx, and a square cross-section (square cross-section, as mentioned in the question) with sides (x^2 - x). The volume of this slice is
(x^2 - x)^2 * dx

(Square face and dx width)

Now, Integrating this between x=2 and x=3, we have

Integral( (x^2 - x)^2 dx, between x=2, x=3)
= [x^5/5 - 2* x^4/4 + x^3/3, between x=2, x=3]
= [3^5/5 - 3^4/2 + 3^3/3 - (2^5/5 - 2^4/2 + 2^3/3)]
= 481/30

Don't know how you got the pi, part, unless, the cross section is not square, but circular. Though, in that case the answer would be 481/60 * pi, I think

Answer 3

integral of pi (R^2-r^2)
integral of pi (x^4-x^2) between x=2 and 3

pi (1/5x^5-1/3x^3) = pi (3^5 /5 -3^3 /3 - 2^5 /5 + 2^3 /3)

pi (241/5-9 -32/5 +8/3) = pi (723/15-135/15-96/15+40/15)

= pi (763-231)/15 = pi 532/15

That's what I got, I'm a little rusty.

Answer 4

Yes, it is right.