The base of a solid is bounded by the curve y=√ x + 1, the x-axis and the line x = 1. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid.
extra credit: what if the cross-sections are parallel to the x-axis?
To find the volume of the solid, we need to use integration. Since the cross sections are squares, we know that the area of each cross section is given by the side length of the square squared.
Let's start by finding an expression for the area of each cross section. The side length of the square at any given x-coordinate will be equal to the height of the curve at that point, which is given by y = √(x + 1).
The area of each cross section, A, is then given by A = (side length)^2 = (√(x + 1))^2 = x + 1.
Now we need to find the limits of integration. We know that the base of the solid is bounded by the curve y = √(x + 1), the x-axis, and the line x = 1. To find the limits of integration, we need to find the values of x where the curve intersects the x-axis and the line x = 1.
Setting y = √(x + 1) equal to 0 gives us:
√(x + 1) = 0
Squaring both sides gives us:
x + 1 = 0
Solving for x, we find x = -1. This is the point where the curve intersects the x-axis.
The line x = 1 represents the rightmost boundary of the solid. Therefore, the limits of integration will be from x = -1 to x = 1.
Now we can set up the integral to calculate the volume:
V = ∫[from -1 to 1] (x + 1) dx
Evaluating this integral will give us the volume of the solid.
this is similar to another one.
I assume you mean y = √(x+1) otherwise the boundary is not closed.
The base of each square is just y. So, since each square patch has thickness dx, the sum of their volumes is just
∫[-1,1] y^2 dx = ∫[-,1] (x+1) dx