Let R be the region bounded by y = ln x, x-axis, and x = 3. Answer the following.

a) Write, but do not solve, an integral expression that will find the volume of the solid that results from rotating R about x = -2.

b) Write, but do not solve, an integral expression that will find the volume of the solid that results from rotating R about y = 4.

I am confused about this question as well. Could you please help me understand how to do this? I would greatly appreciate it!

(a) using shells,

v = ∫[1,3] 2πrh dx
where r = x+2 and h=y=ln(x)

using discs,
v = ∫[0,ln3] π(R^2-r^2) dy
where R=5 and r=x+2=2+e^y

(b) use the same ideas with a horizontal axis. Just keep in mind the volume of a disc (or a washer with a hole inside) and a cylinder. I usually try to do it both ways to make sure the answers agree.

Sure! Let's break down the problem step-by-step:

a) To find the volume of the solid that results from rotating region R about the line x = -2, we can use the method of cylindrical shells.

The integral expression to find the volume can be written as:

V = 2π ∫[f(x) * (x + 2)] dx

Where f(x) represents the height of the shell at a given x-coordinate. In this case, since we are revolving R about x = -2, we use (x + 2) as the radius.

b) To find the volume of the solid that results from rotating region R about the line y = 4, we can use the method of cylindrical shells as well, but with a slight modification.

The integral expression to find the volume can be written as:

V = 2π ∫[x * (g(y) - 4)] dy

Where g(y) represents the inverse function of f(x), which is x = e^y. In this case, we use (g(y) - 4) as the radius, since we are revolving R about y = 4.

To solve these integrals, you can use the appropriate limits of integration for each case and evaluate the integral. I hope this helps! Let me know if you have any further questions.

Sure! I'd be happy to help you understand how to approach these problems.

a) To find the volume of the solid that results from rotating region R about the line x = -2, we can use the disk method. The disk method involves integrating cross-sectional areas of infinitesimally small disks perpendicular to the axis of rotation. In this case, the axis of rotation is x = -2.

To set up the integral expression for the volume, we need to consider the radius of each disk. For any x value within region R, the distance from the axis of rotation (x = -2) to that point is (x - (-2)) = (x + 2).

The height of each disk will be the difference between the values of the curve y = ln x and the x-axis, which is y - 0 = y.

Therefore, the area of each disk will be π(radius)^2, where the radius is (x + 2) and the height is y = ln x.

Now, we need to integrate the areas of all the disks over the interval [a, b], where a and b are the x-coordinates that define the region R. In this case, the region R is bounded by the curve y = ln x, the x-axis, and x = 3. So, the integral expression for the volume can be written as:

V = ∫[a, b] π(radius)^2 dy

Since we are given the equation y = ln x, we need to express the equation in terms of x to determine the limits of integration.

To do this, we solve the equation y = ln x for x:

ln x = y
=> x = e^y

Therefore, the limits of integration become a = 0 (when y = 0, x = 1) and b = ln 3 (when y = ln 3, x = 3). The integral expression for the volume thus becomes:

V = ∫[0, ln 3] π((e^y) + 2)^2 dy

b) To find the volume of the solid that results from rotating region R about the line y = 4, we can use the shell method. The shell method involves integrating cylindrical shells with height dy along the axis of rotation.

Similar to the previous case, we need to consider the radius and height of each cylindrical shell. The radius of each shell will be the difference between the x-coordinate at each point on the curve y = ln x and the line y = 4, i.e., x - 4. The height of each shell will be the difference between the x-coordinates that define the region R, which is 3 - e^y.

Therefore, the volume of each shell will be the product of the circumference (2π(radius)) and the height.

Now, we need to integrate the volumes of all the shells over the interval [c, d], where c and d are the y-coordinates that define the region R. In this case, the region R is bounded by the curve y = ln x, the x-axis, and x = 3. So, the integral expression for the volume can be written as:

V = ∫[c, d] 2π(radius) * height dy

Since we are given the equation y = ln x, the limits of integration will be the same as in part a), i.e., c = 0 and d = ln 3. The integral expression for the volume thus becomes:

V = ∫[0, ln 3] 2π(y - 4)(3 - e^y) dy

I hope this helps clarify the approach to solving these problems! Let me know if you have any further questions.