# Thread: Surface area over y-axis (integrating wrong)

1. ## Surface area over y-axis (integrating wrong)

The book says the answer is

$\displaystyle 15\pi\sqrt{17}$

I have tried to manipulate the numbers to try to get that answer, but I don't know what mistake(s) I am doing.

2. ## Re: Surface area over y-axis (integrating wrong)

Hey Hearts.

You are using the wrong variable if you are rotating about the y-axis - it should be a function of y not a function of x if you are indeed rotating it about the x-axis.

The derivative factor will also have to change meaning you will be looking at dx/dy not dy/dx since the slopes will be different.

If you can state the formula you are using it would be appreciated - otherwise you will have to state how you got it. You are correct in that you will have to take the arc-lengths of each cross section and sum them to get the surface area [which is sort of what the integral is looking to do].

If you took an arc-length approach you should get a lot of circles with varying cross sections that have their origin at the y-axis and you sum them up. If you do the arc-length approach you can introduce an argument of theta from 0 to 2*pi and make that a function of y and use that to get the area.

I'll wait for your response.

3. ## Re: Surface area over y-axis (integrating wrong)

$\displaystyle \sqrt{1+\left(\frac{1}{4}\right)^2}$

$\displaystyle \sqrt{1+4^2}$

4. ## Re: Surface area over y-axis (integrating wrong)

Originally Posted by chiro
Hey Hearts.

You are using the wrong variable if you are rotating about the y-axis - it should be a function of y not a function of x if you are indeed rotating it about the x-axis.

The derivative factor will also have to change meaning you will be looking at dx/dy not dy/dx since the slopes will be different.

If you can state the formula you are using it would be appreciated - otherwise you will have to state how you got it. You are correct in that you will have to take the arc-lengths of each cross section and sum them to get the surface area [which is sort of what the integral is looking to do].

If you took an arc-length approach you should get a lot of circles with varying cross sections that have their origin at the y-axis and you sum them up. If you do the arc-length approach you can introduce an argument of theta from 0 to 2*pi and make that a function of y and use that to get the area.

I'll wait for your response.
Originally Posted by Idea
$\displaystyle \sqrt{1+\left(\frac{1}{4}\right)^2}$

$\displaystyle \sqrt{1+4^2}$