Consider the curve f(x)=x^4 between x = -1 and x = 4.
a)What is the volume obtained by revolving the area under the curve around the x-axis?
b)What is the volume obtained by revolving the area under the curve (i.e. between f(x) and the x-axis) around the line y = -5?
To find the volume obtained by revolving the area under the curve around the x-axis, we can use the method of disks or washers.
First, let's set up the integral for the volume. Since the curve is given by f(x) = x^4, the area between the curve and the x-axis can be expressed as the integral of π(f(x))^2 with respect to x, from x=-1 to x=4:
V1 = ∫[from -1 to 4] π(f(x))^2 dx
Substituting the equation for f(x), we have:
V1 = ∫[from -1 to 4] π(x^4)^2 dx
= ∫[from -1 to 4] πx^8 dx
Now, we can calculate this integral to find the volume:
V1 = π ∫[from -1 to 4] x^8 dx
= π [1/9 x^9] [from -1 to 4]
= π (4^9/9 - (-1)^9/9)
= π (262144/9 + 1/9)
= 262145π/9
Therefore, the volume obtained by revolving the area under the curve around the x-axis is 262145π/9.
Moving on to part b), to find the volume obtained by revolving the area under the curve around the line y = -5, we use the method of shells.
The distance between the curve f(x) = x^4 and the line y = -5 is 5 + f(x), since f(x) is always positive. Therefore, the height of each shell will be 2π(5+f(x)).
As for the width, we can divide the interval from x = -1 to x = 4 into small increments and consider a representative slice at x with width Δx.
The volume of each shell can be approximated as the area of the shell multiplied by the width Δx:
dV = 2π(5+f(x))Δx
To find the total volume, we sum up all these shells by integrating with respect to x:
V2 = ∫[from -1 to 4] 2π(5+f(x)) dx
Substituting f(x) = x^4, we have:
V2 = ∫[from -1 to 4] 2π(5+x^4) dx
Now, we can calculate this integral to find the volume:
V2 = 2π ∫[from -1 to 4] (5+x^4) dx
= 2π [5x + (1/5)x^5] [from -1 to 4]
= 2π [(20/5 + (1/5)(4^5)) - (-5/5 + (1/5)(-1^5))]
= 2π [(20/5 + 1024/5) - (-5/5 - 1/5)]
= 2π [(1044/5) - (-6/5)]
= 2π (1050/5)
= 210π/5
= 42π
Therefore, the volume obtained by revolving the area under the curve (between f(x) and the x-axis) around the line y = -5 is 42π.