The base of a solid in the region bounded by the graphs of y = e^-x, y = 0, and x = 0, and x = 1. Cross sections of the solid perpendicular to the x-axis are semicircles. What is the volume, in cubic units, of the solid?
Answers:
1)(pi/16)e^2
2)(pi/16)(1-1/e^2)
3)(pi/4)(1-1/e^2)
4)(pi/16)(e^2-1)
I'm so confused. I know the volume of a semicircle would be 1/2pir^2h. h would be dx and I think finding the radius is what's throwing me off. Can someone help?
To find the volume of the solid, you need to integrate the areas of the semicircles along the x-axis.
First, let's consider a cross-sectional slice of the solid at a certain x-value. The height of the slice will be given by the function y = e^(-x) since the solid is bounded by y = e^(-x) and y = 0. The radius of the semicircle at this particular x-value will be equal to y, as it is perpendicular to the x-axis.
So, the area of the semicircle at this x-value will be:
A = (1/2) * π * (y)^2
To find the volume, you need to integrate this area over the range of x from 0 to 1. So the integral for the volume will be:
V = ∫[0 to 1] (1/2) * π * (e^(-x))^2 dx
Simplifying this integral, we get:
V = (1/2) * π * ∫[0 to 1] e^(-2x) dx
To solve this integral, we can use the substitution u = -2x, which gives us:
du = -2 dx
dx = (-1/2) du
Substituting back into the integral, we get:
V = (1/2) * π * ∫[0 to -2] e^u * (-1/2) du
V = (-1/4) * π * ∫[0 to -2] e^u du
Now, integrating e^u, we get:
V = (-1/4) * π * [e^u] [from 0 to -2]
V = (-1/4) * π * (e^(-2) - e^0)
Since e^0 = 1, the volume is:
V = (-1/4) * π * (e^(-2) - 1)
To simplify further, we can write e^(-2) as 1/e^2:
V = (-1/4) * π * (1/e^2 - 1)
Multiplying through by -1/4:
V = (π/16) * (1 - 1/e^2)
So, the correct answer is option 2) (π/16) * (1 - 1/e^2).
huh. You seem to have it in hand. The diameter of each semicircle is just y=e^-x, so
v = ∫[0,1] πd^2/8 dx
where d=y=e^-x
v = ∫[0,1] π/8 e^(-2x) dx = π/16 (1 - 1/e^2)
Too bad you couldn't show some of your work ...