Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -3
I got the integral from 1 to 2.718 of pi(1)^2-pi(ln(x))^2
Is this correct or is there a trick I'm missing because that's often the case with calculus
what's this 2.718 stuff? That is e. Use it, just like π, without some sloppy decimal approximation.
since the axis is the line y = -3, the radius of each disc is y+3. So,
v = ∫[1,e] π((1+3)^2-(lnx + 3)^2) dx
Your setup of the integral is partially correct, but there is a small mistake. Let's go through the steps to correctly set up the integral.
To find the volume of the solid formed by revolving the region bounded by the curves y = Ln(x), y = 1, and x = 1 around the line y = -3, we can use the method of cylindrical shells.
1. First, sketch the graphs of the curves y = Ln(x), y = 1, and x = 1. The region bounded by these curves lies between x = 1 and x = e (approximately 2.718).
2. Now, consider a vertical strip at a particular x-value, say x. Imagine rotating this strip around the line y = -3 to form a cylindrical shell.
3. The height of the cylindrical shell is given by y = (Ln(x) - 1), as it represents the difference in y-values between the curves Ln(x) and 1.
4. The radius of the cylindrical shell is the distance from the line y = -3 to the curve y = Ln(x). Since the line y = -3 is a horizontal line, the distance is simply (-3 - Ln(x)).
5. The differential width of the strip is dx.
6. The volume of the cylindrical shell is given by the formula:
V = 2π(radius) * (height) * (width)
= 2π(-3 - Ln(x)) * (Ln(x) - 1) * dx
7. To find the total volume, we integrate the expression for the volume of each cylindrical shell over the interval from x = 1 to x = e:
∫[1 to e] 2π(-3 - Ln(x)) * (Ln(x) - 1) dx
So, the correct integral to set up the volume is:
∫[1 to e] 2π(-3 - Ln(x)) * (Ln(x) - 1) dx
Now you can proceed to evaluate this integral to find the volume.