Did I make a mistake in this integral?

[Dethrone]

I'm really exhausted mentally, so it'll be really helpful if someone can tell me where I made a mistake. I'm rotating surfaces, and with that, I had to solve this integral:

$$=\pi \int \sqrt{64-3x^2} dx$$
$$=\frac{\pi}{\sqrt{3}} \int \sqrt{\frac{64}{3}-x^2} dx$$

Let $$x = \frac{8}{\sqrt{3}} sin\left({\theta}\right) d\theta$$
$$dx = \frac{8}{\sqrt{3}} cos\left({\theta}\right) d\theta$$

=$$\frac{8\pi }{3} \int \cos^2\left({\theta}\right) (\frac{8}{\sqrt{3}}) d\theta$$

Have I made a mistake already?

 

[MarkFL]

As a quick first thought, are you using the disk method, and as such, should the integrand be squared?

 

[Dethrone]

I'm calculating surface area, so I'm rotating the line across. The problem basically boils down to what I have above, and I know I'm right up to this point. I know I messed up this integral somehow, but I'm so tired that I don't know how. I've done hundreds of trig. substitution questions, not sure what I'm stuck on this one. My answer is off by a factor of 3, so I want to check my process.

 

[Dethrone]

I guess I could type out more steps, but my brain isn't cooperating with me.

After:

$$=\frac{64 \pi}{3\sqrt{3}} \int \frac{1}{2} (1+\cos\left({2\theta}\right) d\theta$$
$$=\frac{64 \pi}{6\sqrt{3}} (\theta + \frac{1}{2} \sin\left({2\theta}\right)) $$
$$=\frac{64 \pi}{6\sqrt{3}} (\theta + \sin\left({\theta}\right)\cos\left({\theta}\right)) $$

The original question is "revolve $\frac{x^2}{16} + \frac{y^2}{4} = 1$ about the x-axis", but I've left out the bounds, etc because my mistake is the integral.

$$S=2 \pi \int_{-4}^{4} y\sqrt{1+(\d{y}{x})^2} \,dx $$

 

[MarkFL]

Ah, yes, I see you stated you are finding a surface of rotation...sorry for not reading thoroughly. :D

So, you begin with:

\(\displaystyle I=\pi\int\sqrt{64-3x^2}\,dx\)

I would let:

\(\displaystyle x=\frac{8}{\sqrt{3}}\sin(\theta)\,\therefore\,dx=\frac{8}{\sqrt{3}}\cos(\theta)\,d\theta\)

And so we now have:

\(\displaystyle I=\frac{64\pi}{\sqrt{3}}\int\cos^2(\theta)\,d\theta\)

This is indeed 3 times greater than what you stated. Do you see your error in the first step you made in the first post?

 

[Dethrone]

My head is so fried right now, but I found my mistake.

It should be this:
$$\sqrt{3(\frac{64}{3}-x^2})=\sqrt{3}\sqrt{\frac{64}{3}-x^2}$$

I was trying to factor out 3, but instead, I did something quite careless...(Rofl) I knew it was a dumb mistake from the start...(Crying)