# Surface area of a curve

1. Feb 27, 2013

### stunner5000pt

1. The problem statement, all variables and given/known data
The given curve is rotated about the y-axis. Find the area of the resulting surface.
$$y = 3 - x^2$$
$0 \leq x \leq 4$

2. Relevant equations
$$A_{y} =2\pi\int_a^b x \sqrt{1+\left(\frac{dx}{dy}\right)^2} dy$$

3. The attempt at a solution
now we need to write x in terms of y so we get:
$$x = \pm \sqrt{3-y}$$
and if we differentiate the above, we get:

$$\frac{dx}{dy} = -\frac{1}{\sqrt{3-y}}$$

Our range also changes
when x = 0, then y = 3
and x = 4, then y = -13

so then our integral is:

$$A_{y} =2\pi\int_{-13}^{3} 2\sqrt{3-x} \sqrt{1+ \left(-\frac{1}{\sqrt{3-y}}\right)^2} dy$$

So i multiplied the above by 2 instead of plus or minus
and in addition to that, do we need to consider the dx/dy term? Does that need to be multiplied as well?

2. Feb 27, 2013

### Dick

For one thing, I think your derivative is a bit off. For another thing if x=sqrt(3-y) why did change it to sqrt(3-x) in your integral? And you don't really have to worry about the +/- bit. x in your integral is really a radius. The sign doesn't matter. Draw a picture.

Last edited: Feb 27, 2013
3. Feb 27, 2013

### haruspex

It might have been a little easier to work with ds =√(1+(dy/dx)2).dx instead, but it should work either way.
I don't understand why you multiplied by 2 at the end. The choice between ±√ for the value of x is just that, a choice. One will give you a negative result, one positive. You know the answer is to be positive, so just take the positive value at the end.

4. Feb 27, 2013

### stunner5000pt

Thanks for your replies. Let me fix what you have mentioned:

The derivative is off:

$$x = \pm \sqrt{3-y}$$

$$\frac{dx}{dy} = -\frac{1}{2 \sqrt{3-y}}$$

And fixing that and the 2 in the integral:
$$A_{y} =2\pi\int_{-13}^{3} \sqrt{3-y} \sqrt{1+ \left(\frac{1}{2 \sqrt{3-y}}\right)^2} dy$$

How does this look now?

5. Feb 27, 2013

### haruspex

Looks good now.

6. Feb 27, 2013

### Dick

Much better. Now if you can try to work it out without Wolfram, that would be better yet. It's really not terribly hard.

7. Feb 27, 2013

### stunner5000pt

since they are both under root we can multiply the 3-y into the big root. this will yield under the root $-y +\frac{13}{4}$ which we can integrate by guessing or by u-substitution.

Thank you!!

8. Feb 22, 2015

### Shelby Roy

Can someone explain the range change to me?

9. Feb 23, 2015

### haruspex

The original range was stated as bounds on x. Because the integral was constructed as with respect to y, the solver had to find the range for y. As I remarked at the time, the whole problem would have been easier using x as the variable of integration.