The region bounded by y=3/(1+x^2), y=0, x=0 and x=3 is rotated about the line x=3. Using cylindrical shells, set up an integral for the volume of the resulting solid. The limits of integration are:
v = ∫[0,3] 2πrh dx
where r=3-x and h=3/(1+x^2)
or, using discs,
v = ∫[0,3] π(R^2-r^2) dy
where R=3 and r=3-√(3/y - 1)
To set up the integral using cylindrical shells, we need to express the volume of each cylindrical shell in terms of its radius and height.
Let's consider an infinitesimally thin vertical strip with width Δx and height y. When this strip is rotated about the line x=3, it traces out a cylindrical shell with radius (3-x) and height Δy (which approximates Δx as it becomes infinitesimally small).
The volume of each cylindrical shell is given by the formula:
dV = 2π(3-x)⋅y⋅Δx
Now, we need to find an expression for y in terms of x. The equation of the curve y = 3/(1 + x^2) describes the top boundary of the region.
To find the limits of integration, we need to determine the values of x where the strip crosses the x-axis, as this corresponds to the limits of the region.
Since y = 0 represents the x-axis, we need to solve the equation:
0 = 3/(1 + x^2)
Multiplying both sides by (1 + x^2), we get:
0 = 3
Since there is no x value that makes this equation true (3 is never equal to 0), the region does not cross the x-axis (y = 0) within the given interval. Therefore, the limits of integration for x are from 0 to 3.
Now, we can set up the integral for the volume:
V = ∫[0 to 3] 2π(3-x)⋅y⋅dx
Substituting y = 3/(1 + x^2), we have:
V = ∫[0 to 3] 2π(3-x)⋅(3/(1 + x^2))⋅dx
These are the limits of integration for the cylindrical shell method when using the region bounded by y=3/(1+x^2), y=0, x=0, and x=3 and rotating about the line x=3.
To find the volume of the solid using cylindrical shells, we need to set up an integral that represents the sum of the volumes of all the infinitesimally thin cylindrical shells.
The formula for the volume of a cylindrical shell is given by V = 2πrhΔx, where:
- V is the volume of the shell,
- π is a constant (approximately 3.14159),
- r is the distance from the axis of rotation to the shell (in this case, r = x - 3 since the axis of rotation is the line x = 3),
- h is the height of the shell (equal to y), and
- Δx is the thickness of the shell.
To set up the integral, we need to find the limits of integration for x. Let's examine the regions defined by the intersection of the curves y = 3/(1 + x^2) and y = 0.
When y = 3/(1 + x^2), we have:
3/(1 + x^2) = 0
This equation has no real solutions since 3 divided by any positive number will always be greater than zero. Therefore, the upper limit of integration for x is at the point where y = 0, which occurs at x = 3.
The lower limit of integration for x is at the point where the curve intersects the y-axis, which is at x = 0.
So, the limits of integration for x are from x = 0 to x = 3.
Therefore, the integral for the volume of the solid using cylindrical shells is given by:
V = ∫[from 0 to 3] 2π (x - 3) (3/(1 + x^2)) dx
Simplifying and combining terms, we have:
V = 2π ∫[from 0 to 3] (3(x - 3))/(1 + x^2) dx
This integral represents the volume of the solid obtained by rotating the region bounded by y = 3/(1 + x^2), y = 0, x = 0, and x = 3 about the line x = 3 using cylindrical shells.