Volume of Solid using cross sections

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To find the volume of the solid with a base bounded by the curves y=x+1 and y=x^2-1, the area of the square cross-section is determined by the difference between these two functions. The correct setup for the integral involves calculating the square of the difference of the curves, leading to the expression for volume as the integral from -1 to 2 of ((x+1)-(x^2-1))^2 dx. The side length of the square at any x-value is given by this difference, and the area of the square is then the square of this side length. The discussion emphasizes understanding the correct setup for the integral to solve for the volume effectively.
BigJon
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Homework Statement


Find the volume of a solid whose base bounded by y=x+1 and y=x^2-1, with cross section of a square perpendicular to x-axis


The Attempt at a Solution


So i set up the problem like this delta volume= y^2*delta x, that being the area of the square where i get loss is that I am supposed to subtract the curves from one another but don't know if its setup like this:

integral from -1 to 2 ((x+1)-(x^2-1))^2 dx or ((x+1)^2)-((x^2-1)^2) dx, from here i know how to do the rest
 
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What do you think the length of the side of the square is for a given value of x? Then what would be the area of that square?
 
Haha thanx, now that you put it that way :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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