Find the volume of the solid whose base is bounded by the graphs of y= x + 1 and y= x^21 with square cross sections taken perpendicular to the xaxis.
Thanks!
Thanks!

The curves intersect at x = 1 and x = 2. Since the cross sections are squares, all we require is a formula for one side of the square as a function of x (squared of course, to get the area.)
V = ∫ A(x) dx from 1 to 2
Now, the base of the square is given by the distance between the two curves. Since the line is above the parabola, subtract the parabola from the line.
[x + 1]  [x^2  1]
 x^2 + 1 + x + 1
 x^2 + x + 2
But then, the area of a square will be that expression squared:
[ x^2 + x + 2][ x^2 + x + 2] = x^4  2 x^3  3 x^2 + 4 x + 4
A(x) = x^4  2 x^3  3 x^2 + 4 x + 4
Integrate that up now:
V = ∫ [x^4  2 x^3  3 x^2 + 4 x + 4] dx from 1 to 2
V = [(1/5)x^5  (1/2)x^4  x^3 + 2x^2 + 4x] from 1 to 2
Evaluate:
V = [(1/5)(2)^5  (1/2)(2)^4  (2)^3 + 2(2)^2 + 4(2)]  [(1/5)(1)^5  (1/2)(1)^4  (1)^3 + 2(1)^2 + 4(1)]
V = 81/10
V = 8.1
Done!
Edit: Fixed a transcription error; answer remains unchanged.
V = ∫ A(x) dx from 1 to 2
Now, the base of the square is given by the distance between the two curves. Since the line is above the parabola, subtract the parabola from the line.
[x + 1]  [x^2  1]
 x^2 + 1 + x + 1
 x^2 + x + 2
But then, the area of a square will be that expression squared:
[ x^2 + x + 2][ x^2 + x + 2] = x^4  2 x^3  3 x^2 + 4 x + 4
A(x) = x^4  2 x^3  3 x^2 + 4 x + 4
Integrate that up now:
V = ∫ [x^4  2 x^3  3 x^2 + 4 x + 4] dx from 1 to 2
V = [(1/5)x^5  (1/2)x^4  x^3 + 2x^2 + 4x] from 1 to 2
Evaluate:
V = [(1/5)(2)^5  (1/2)(2)^4  (2)^3 + 2(2)^2 + 4(2)]  [(1/5)(1)^5  (1/2)(1)^4  (1)^3 + 2(1)^2 + 4(1)]
V = 81/10
V = 8.1
Done!
Edit: Fixed a transcription error; answer remains unchanged.