Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^3,y=0,x=4,x=7. About the y-axis.
Volume =
v = ∫[4,7] 2πrh dx
where r=x and h=y=1/x^3
or,
v = ∫[1/343,1/64] π(R^2-r^2) dy
where r=4 and R=1/&$x221B;y
that's R=1/cbrt(y)
∛
To find the volume of the solid obtained by rotating the region bounded by the given curves about the y-axis, we can use the method of cylindrical shells.
First, let's sketch the graph of the region bounded by the curves y=1/x^3, y=0, x=4, and x=7.
Next, we'll imagine slicing the region into thin vertical strips or shells. Each shell will have a height of Δy and a radius equal to the distance from the y-axis to the curve y=1/x^3.
To determine the radius, we need to solve for x in terms of y, using the equation y=1/x^3. Rearranging the equation, we find x = (1/y)^(1/3).
Now, we can express the volume of each shell as Vshell = 2π(radius)(height). Plugging in the values, we have Vshell = 2π((1/y)^(1/3))(Δy).
To determine the limits of integration for y, we look at the bounds on y values in the region. In this case, y ranges from 0 to the value where the two curves intersect. To find this value, we set the equations equal to each other: 1/x^3 = 0, which only occurs at x = ∞. Therefore, the upper limit for y is ∞.
To set up the integral, we need to express the volume V as a sum of all the individual shell volumes by integrating. Taking the integral over the range of y values, we have:
V = ∫(0 to ∞) 2π((1/y)^(1/3))(Δy)
Simplifying, we have:
V = 2π ∫((1/y)^(1/3))(Δy)
Now, we find the antiderivative of the integrand and evaluate the integral:
V = 2π [(-3/2)(1/y)^(2/3)] (evaluated from 0 to ∞)
Plugging in the limits of integration, we have:
V = 2π [0 - (-3/2)(1/∞)^(2/3)]
As (∞)^(2/3) approaches infinity, the second term in brackets becomes zero.
Therefore, the volume V is equal to 2π times zero, which is zero.
Therefore, the volume of the solid obtained by rotating the region bounded by the given curves about the y-axis is zero.