Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis.
y=ln x, y=0, x=1, x=3
Please show the steps in your work, thanks! :)
The differential volume is from y=ln1 to y=ln3 from x=1 to 3
dv= 2PI y dx= 2PI ln x dx
V= INT 2PI ln x dx from x=1 to 3
V= 2PI (xlnx -x) from ...
To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis, we can use the method of cylindrical shells.
First, let's sketch the region bounded by the graphs of the equations y = ln x, y = 0, x = 1, x = 3.
This region is a finite area that lies between the x-axis and the curve y = ln x. It extends from x = 1 to x = 3, and it is bounded by the y-axis and the curve y = 0.
Next, we need to express the functions in terms of x, as the cylindrical shell method requires integration with respect to x.
We already have y = 0, which remains the same.
For y = ln x, we can rewrite it as x = e^y, since ln x and e^x are inverse functions.
Now, let's consider a thin vertical strip within this region, centered at x with a width dx (infinitesimally small width).
The height of this strip is given by the difference between the two functions: h(x) = ln x - 0 = ln x.
The circumference of the cylindrical shell, which we will revolve around the x-axis, is given by the formula 2πr, where r is the distance from the x-axis to the strip. In this case, r is equal to x.
The volume of this cylindrical shell is given by V = 2πrh(x)dx = 2πx(ln x)dx.
To find the total volume, we need to integrate this expression over the interval from x = 1 to x = 3, as those are the bounds of the region.
∫(1 to 3) [2πx(ln x)] dx
Now, we can evaluate this integral:
= 2π ∫(1 to 3) (x ln x) dx
To calculate this integral, we can use integration by parts. Let u = ln x, and dv = x dx.
Taking the derivative of u, we get du = (1/x) dx.
Since dv = x dx, integrating dv gives us v = (1/2)x^2.
Now, we can use the formula for integration by parts:
∫ u dv = uv - ∫ v du
Plugging in the values, we have:
= 2π [(1/2)(x^2 ln x) - ∫ (1/2)x^2 (1/x) dx]
= π [(x^2 ln x) - (1/2) ∫ x dx]
= π [(x^2 ln x) - (1/2)(x^2/2)]
= π [x^2 ln x - x^2/4] evaluated from 1 to 3
Evaluating the expression at x = 3, we get:
= π [(3^2 ln 3) - (3^2/4)]
Evaluating the expression at x = 1, we get:
= π [(1^2 ln 1) - (1^2/4)]
Since ln 1 is equal to 0 and 1^2 = 1, we can simplify this to:
= π [(9 ln 3) - (9/4)]
And finally, we can compute the value:
≈ π (9 ln 3 - 9/4) cubic units
So, the volume of the solid generated by revolving the region bounded by the graphs of y = ln x, y = 0, x = 1, and x = 3 about the x-axis is approximately equal to π (9 ln 3 - 9/4) cubic units.