Find the volume if the area between the curve y=lnx, x=1, and y=1 is rotated about the x-axis.
I tried the integral from 1 to e of 1-lnx^2 but that doesn't work.
You want
pi{integral)(1-lnx)^2 dx from 1 to e
To find the volume of the solid formed by rotating the area between the curve y = ln(x), x = 1, and y = 1 about the x-axis, we can use the method of cylindrical shells.
Step 1: Determine the limits of integration.
Since we are rotating about the x-axis, the limits of integration will be the x-values where the curves intersect. We need to find the intersection points between y = ln(x) and y = 1.
Setting ln(x) = 1, we have:
ln(x) = 1
x = e^1
x = e
So the limits of integration will be x = 1 and x = e.
Step 2: Determine the height of each cylindrical shell.
The height of each cylindrical shell will be the difference between the y-values of the curves at a given x-value. In this case, it will be 1 - ln(x).
Step 3: Determine the radius of each cylindrical shell.
The radius of each cylindrical shell will be the x-value at a given point on the curve.
Step 4: Set up the integral for the volume.
The volume of the solid can be found by integrating the product of the height and the circumference of each cylindrical shell. The circumference of a cylindrical shell is given by 2πr, where r is the radius of the shell.
So the integral for the volume is:
V = ∫[1 to e] 2πx(1 - ln(x)) dx
Step 5: Evaluate the integral.
This integral can be evaluated using techniques of integration. Integrating the function gives:
V = ∫[1 to e] 2πx - 2πxln(x) dx
= [πx^2 - 2πxln(x)] evaluated from x = 1 to x = e
= (πe^2 - 2πe - π + 2π)
Therefore, the volume of the solid formed by rotating the area between the curve y = ln(x), x = 1, and y = 1 about the x-axis is (πe^2 - 2πe - π + 2π) cubic units.
To find the volume of the solid obtained by rotating the area between the curve y = ln(x), the vertical line x = 1, and the horizontal line y = 1 about the x-axis, you can apply the method of cylindrical shells.
The volume of the solid can be determined by integrating the circumference of infinitesimally thin cylindrical shells along the interval [1, e] (where e is the base of the natural logarithm) and multiplying it by the height (or thickness) of each shell. Since the shells are infinitesimally thin, the height is given by dx, the infinitesimal change in x.
Now, let's calculate the circumference of each shell. The circumference of a shell is given by 2πr, where r is the distance from the axis of rotation (in this case, the x-axis) to the curve y = ln(x). Since the axis of rotation is the x-axis, the distance is y = ln(x).
Moreover, the height of each shell is dx, the infinitesimal change in x.
So, the volume of each shell is given by dV = 2π(ln(x))dx.
To find the total volume, you need to integrate this expression from x = 1 to x = e:
V = ∫[1, e] 2π(ln(x))dx.
Now, we can evaluate this integral:
V = 2π∫[1, e] ln(x)dx.
Using integration by parts (or integration tables for logarithmic functions), you can find that:
∫ ln(x)dx = x(ln(x) - 1) + C,
where C is the constant of integration.
Therefore,
V = 2π[x(ln(x) - 1)] from 1 to e.
Substituting the limits of integration, we get:
V = 2π[e(ln(e) - 1)] - 2π[1(ln(1) - 1)].
Since ln(e) = 1 and ln(1) = 0, this simplifies to:
V = 2π[e - 1].
Hence, the volume of the solid obtained by rotating the area between the curve y = ln(x), x = 1, and y = 1 about the x-axis is 2π(e - 1) cubic units.