Calculus: Cross Section Problem, Finding Volume
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Calculus: Cross Section Problem, Finding Volume

[From: ] [author: ] [Date: 11-06-08] [Hit: ]
all we require is a formula for one side of the square as a function of x (squared of course, to get the area.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.......
Find the volume of the solid whose base is bounded by the graphs of y= x + 1 and y= x^2-1 with square cross sections taken perpendicular to the x-axis.

Thanks!

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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.
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