MHB Surface Area of an Ellipse Obtained by Rotation

[renyikouniao]

A surface is obtained by rotating around the x-axis the arc over the integral(-1,0.5) of an ellipse given by:
x^2+4y^2=1

What is its surface area?

Here's my solution:

I use the equation:
S=integral( upper bound: a lower bound: b ) 2(pi)y*[1+(f'(x))^2]^0.5 dx

Since x^2+4y^2=1
y=[(1-x^2)/4]^0.5
dy/dx=-x/[2(1-x^2)^0.5]

S=integral (upper bound: 0.5 lower bound: -1) 2(pi)[(1-x^2)/4]^0.5 * [1 -x/[2(1-x^2)^0.5]]^0.5

And I have no idea how to evaluate this whole thing...Am I right so far?

 

[MarkFL]

Re: Surface area

I find (using the top-half of the ellipse):

$$f(x)=\frac{1}{2}\sqrt{1-x^2}$$

$$f'(x)=-\frac{x}{2\sqrt{1-x^2}}$$

and so the surface of rotation is given by:

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

$$S=\frac{\pi}{2}\int_{-1}^{\frac{1}{2}}\sqrt{4-3x^2}\,dx$$

At this point, I suggest the substitution:

$$x=\frac{2}{\sqrt{3}}\sin(\theta)$$

Can you proceed?