# Surface area obtained by solid of revolution

• lillybeans
In summary, the problem is that the student is not able to solve the problem if the axis of rotation is shifted. They need to find the surface area of the solid of revolution if the axis is shifted by 4 units.
lillybeans

## Homework Statement

Find surface area of solid of revolution obtained by rotating the curve:

y=x2/40-5lnx

from x=5 to x=7, rotated about x=-4

## The Attempt at a Solution

The problem is I know how to do this if I rotated it about x-axis/y-axis, but I have no idea how to do it if the axis is shifted. I did it pretending it was rotated about the y-axis (since x=-4 is parallel to the y-axis), and my answer came out wrong.

I used:

SA=∫2∏x ds, where ds=√(1+[f'(x)]2) dx

So I found the derivative of the function, everything was fine. I squared and square-rooted, found the integral to be:

∏/10 ∫ x^2+100 dx (evaluated from 5 to 7)

but i didn't use the piece of information "axis of rotation is x=-4" anywhere in my solution. I don't really know how to go about using it, please help?

Do I just change the formula from SA=∫2∏x ds to SA=∫2∏(x+4) ds since the axis is shifted to the left 4 units, so the distance to the axis of rotation would be 4 units longer?

Although the answer is still not correct...

or do I change the function so that it goes from y=x^2/40-5lnx to y=(x-4)^2/40-5ln(x-4)? Or none of the above?:/

Last edited:
Deleted.

Last edited:
BluFoot said:
You are correct in that the radius is equal to (x+4).

The volume should be equal to:

∫ from 5 to 7 2π(x+4)(x^2/40-5lnx) dx

Is that what you did?

Thanks for answering, but I don't understand the second part of the integral. Why is it (x^2/40-5lnx) dx?

I thought it should be:

∫2∏(x+4) ds

where ds is

√(1+[f'(x)]^2) dx, and f'(x) is the derivative of x^2/40-5lnx?

Deleted.

Last edited:
OHHH NEVERMIND! I'm stupid, you're looking for surface area, not volume. Ignore everything I've said, sorry!

## 1. What is a solid of revolution?

A solid of revolution is a three-dimensional object that is formed by rotating a two-dimensional shape around an axis. This shape can be any curve or polygon, and the resulting solid will have a symmetrical, curved surface.

## 2. How is the surface area of a solid of revolution calculated?

The surface area of a solid of revolution can be calculated by using the formula 2π∫abf(x)√(1+(f'(x))2) dx, where a and b are the limits of the integral and f(x) is the equation of the curve being rotated.

## 3. What is the relationship between the shape of the curve and the resulting surface area?

The shape of the curve being rotated directly affects the resulting surface area. For example, a curve with a smaller radius will result in a smaller surface area, while a curve with a larger radius will result in a larger surface area.

## 4. Can the surface area of a solid of revolution be approximated?

Yes, the surface area can be approximated by using numerical methods such as the trapezoidal rule or Simpson's rule. These methods involve dividing the curve into smaller sections and calculating the surface area of each section, then summing them together to get an approximate value.

## 5. What real-life applications does the concept of surface area obtained by solid of revolution have?

The concept of surface area obtained by solid of revolution has various real-life applications, such as in engineering and architecture for designing structures like bridges and buildings. It is also used in physics for calculating the moment of inertia of objects. Additionally, it is used in calculus to solve problems involving volumes and surface areas.

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