How Do You Calculate the Surface Area of a Rotated Curve?

[stunner5000pt]

Homework Statement

The given curve is rotated about the y-axis. Find the area of the resulting surface.
$$y = 3 - x^2$$
$0 \leq x \leq 4$

Homework Equations

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

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?

Thanks for your help!

 

[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.

 

[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.

 

[stunner5000pt]

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.

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.

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?

 

[haruspex]

Looks good now.

 

[Dick]

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?

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

 

[stunner5000pt]

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

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!

 

[Shelby Roy]

Can someone explain the range change to me?

 

[haruspex]

Can someone explain the range change to me?

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.