# Can't integrate the surface area of revolving curve the normal way

• aleksbooker
In summary: CHence\int_0^{\ln 2} e^{2y}\,dy = \frac 12 e^{-2y}\bigg|_0^{\ln 2} = \frac 12 (e^{\ln 2} - e^0) = \frac 12 (2 - 1) = \frac 12In summary, the area of the surface generated by revolving the curve x=\frac{e^y + e^{-y} }{2} from 0 \leq y \leq ln(2) about the y-axis can be found by first using the formula S = \int_0^{\ln 2} 2\pi
aleksbooker

## Homework Statement

Find the area of the surface generated by revolving the curve

$x=\frac{e^y + e^{-y} }{2}$

from 0 $\leq$ y $\leq$ ln(2) about the y-axis.

## The Attempt at a Solution

I tried the normal route first...

g(y) = x = $\frac{1}{2} (e^y + e^{-y})$
g'(y) = dx/dy = $\frac{1}{2} (e^y - e^{-y})$

S = $\int 2\pi \frac{1}{2} (e^y + e^{-y}) \sqrt{1+(e^y - e^{-y})^2} dy$

S = $\pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4}e^{2y}+\frac{1}{2}+\frac{1}{4}e^{-2y} } dy$

S = $\pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4} (e^y+e^{-y})^2} dy$

S = $\pi \int (e^y + e^{-y}) \frac {1}{2} (e^y+e^{-y}) dy$

But then I got stuck here...

S = $\frac{1}{2} \pi \int (e^y + e^{-y})^2 dy$

How should I proceed? Thanks in advance.

aleksbooker said:

## Homework Statement

Find the area of the surface generated by revolving the curve

$x=\frac{e^y + e^{-y} }{2}$

from 0 $\leq$ y $\leq$ ln(2) about the y-axis.

## The Attempt at a Solution

I tried the normal route first...

g(y) = x = $\frac{1}{2} (e^y + e^{-y})$
g'(y) = dx/dy = $\frac{1}{2} (e^y - e^{-y})$

S = $\int 2\pi \frac{1}{2} (e^y + e^{-y}) \sqrt{1+(e^y - e^{-y})^2} dy$

I think this should be
$$S = \int_0^{\ln 2} 2\pi \frac 12 (e^y + e^{-y}) \sqrt{1 + \frac14 (e^y - e^{-y})^2}\,dy$$

S = $\pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4}e^{2y}+\frac{1}{2}+\frac{1}{4}e^{-2y} } dy$

And now missing factor of 1/4 has appeared (but the limits are still missing).

S = $\pi \int (e^y + e^{-y}) \sqrt{\frac{1}{4} (e^y+e^{-y})^2} dy$

S = $\pi \int (e^y + e^{-y}) \frac {1}{2} (e^y+e^{-y}) dy$But then I got stuck here...

S = $\frac{1}{2} \pi \int (e^y + e^{-y})^2 dy$

How should I proceed? Thanks in advance.

$(e^y + e^{-y})^2 = e^{2y} + 2 + e^{-2y}$...

I forgot how much I hate working with e. Thanks for pointing out how simple that really was.

Here's what I was doing:

S = $\frac{1}{2}\pi \int (e^{2y} + 2 + e^{-2y}) dy$

Then, breaking it up into three separate integrals and working with just the first one...

S = $\frac{1}{2}\pi \int (e^{2y})$

S = $\frac{1}{2}\pi \frac{1}{2y+1} e^{2y+1}$ from 0 to ln(2)

aleksbooker said:

I forgot how much I hate working with e. Thanks for pointing out how simple that really was.

Here's what I was doing:

S = $\frac{1}{2}\pi \int (e^{2y} + 2 + e^{-2y}) dy$

Then, breaking it up into three separate integrals and working with just the first one...

S = $\frac{1}{2}\pi \int (e^{2y})$

S = $\frac{1}{2}\pi \frac{1}{2y+1} e^{2y+1}$ from 0 to ln(2)

No.
$$\int e^{ay}\,dy = \frac{e^{ay}}a$$

## 1. Why can't I integrate the surface area of a revolving curve using the normal method?

The normal method of integrating the surface area of a curve relies on using known formulas and techniques, such as the Riemann sum or the trapezoidal rule. However, these techniques are only applicable for simple, well-behaved curves. When dealing with more complex curves, such as those formed by revolving a function around an axis, the normal method becomes insufficient and other methods must be used.

## 2. What other methods can be used to integrate the surface area of a revolving curve?

There are several methods that can be used to integrate the surface area of a revolving curve, depending on the specific curve and the level of precision required. Some common methods include using the disk method, the shell method, or the Pappus's centroid theorem. These methods take into account the unique characteristics of a revolving curve and can provide more accurate results.

## 3. How does the disk method work for integrating the surface area of a revolving curve?

The disk method involves slicing the revolving curve into infinitely thin disks, calculating the surface area of each disk, and then adding up the surface areas of all the disks using an integral. This method is often used for revolving curves with a known function and a simple axis of rotation, such as a circle or a parabola.

## 4. Can the shell method be used for any type of revolving curve?

The shell method can be used for many types of revolving curves, including those with a more complex shape or multiple axes of rotation. This method involves slicing the curve into infinitely thin shells, calculating the surface area of each shell, and then adding up the surface areas using an integral. It is often used for curves that cannot be easily solved using the disk method.

## 5. How does Pappus's centroid theorem help with integrating the surface area of a revolving curve?

Pappus's centroid theorem states that the surface area of a solid formed by revolving a curve around an axis is equal to the product of the curve's arc length and the distance traveled by the centroid of the curve. This theorem can be used to calculate the surface area without having to use integrals, making it a useful alternative for curves with a known centroid and axis of rotation.

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