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Find the volume of the solid whose base is bounded by the graphs of \( y=x+1 \) and \( Y=x^{2}-1 \) with the indicated cross-sections taken perpendicular to the \( \mathrm{x} \)-axis. a. square b. rectangle of height 1

User Sudeepta
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To determine the volume of the described solid, we first need to define the bounds of integration by finding the intersection points between the given functions. We educate ourselves that the intersection points are the solutions of the equation \(y1 = y2\). So,

x + 1 = x^2 - 1,

which simplifies to

x^2 - x - 2 = 0;

This equation is a quadratic equation, in the form \(ax^2 + bx + c = 0\) and it can be solved by using the quadratic formula, \(x = (-b ± √(b^2 - 4ac)) / (2a)\):

x = [1 ± sqrt((-1)^2 - 4*1*(-2))] / 2*1,
x = [1 ± sqrt(1 + 8)] / 2,
x = [1 ± sqrt(9)] / 2,
x = [1 ± 3] / 2,

which result in two solutions,

x1 = (1 - 3)/2 = -1, and
x2 = (1 + 3)/2 = 2

For a square cross section, the side length of the square is defined by the difference of y values, \( y2 - y1 = x^2 - 1 - (x + 1) = x^2 - x - 2\). The area of the square is therefore \( (x^2 - x - 2)² \).

To find the volume, integrate the area over the defined interval:

\[\ V_{square} = ∫_{-1}^{2} (x^2 -x -2)^2 dx.\]

The integral is a polynomial that can be directly integrated.

For a rectangular cross section with height 1, the area of the rectangle is \( y2 - y1 = x^2 - x - 2\).

The volume can be found as:

\[ V_{rectangle} = ∫_{-1}^{2}(x^2 -x - 2) dx.\]

Again, this integral is a polynomial and can be directly integrated.

Carrying out these integrations should give the volumes of the solids with the given cross sections.

User Kdb
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