Confused as to how this was simplified.

The question was to find the surface area of a sphere or radius R. I got a bit stuck so I looked at the solution, however I'm confused as to how the simplification in it was attained. The solution is as follows:

$y = \sqrt{R^2 - x^2}$ in the range $-R \leq x \leq R$

$y' = - \frac{x}{\sqrt{R^2 - x^2}}$

Therefore,

$A = 2\pi \int^R_{-R} \sqrt{R^2 - x^2} \left[\frac{x}{\sqrt{R^2 - x^2}}\right]^{1/2} dx$

Then it's somehow simplified to get,

$A = 2\pi \int^R_{-R} \sqrt{R^2 - x^2} \left[\frac{R^2}{\sqrt{R^2 - x^2}}\right]^{1/2}$

Then so on until the final answer is $4 \pi R^2$

But I have no idea how $\frac{x}{\sqrt{R^2 - x^2}}$ became $\frac{R^2}{\sqrt{R^2 - x^2}}$

Any help please?

Since no one's replied. . .
I don't know all that much about this stuff, but should that not be:
$\displaystyle dA=2 \pi x ds = 2 \pi x \sqrt{1+ \left(\dfrac{dy}{dx} \right)^2} \ dx[br]\Rightarrow A = 2\pi \displaystyle\int^R_{-R}x \sqrt{1+ \left(\dfrac{dy}{dx} \right)^2} \ dx$
Then subbing in for $\dfrac{dy}{dx}$
$\Rightarrow A= 2\pi \displaystyle\int^R_{-R} x \sqrt{1+ \dfrac{x^2}{R^2-x^2}} \ dx$
This can then be simplified to the form you were asking about by getting a common denominator in the square root.

The form you've given looks much more like:
$dA=2 \pi y \sqrt{1+ \left(\dfrac{dx}{dy} \right)^2} \ dy$ except that you've subbed in for y and have $\dfrac{dy}{dx}$ instead.
Again, I only touched on this very briefly a while back, but those are my thoughts.

I think $2 \pi x dx$ is for when the function y(x) is a constant.

Therefore,

$A = 2\pi \int^R_{-R} \sqrt{R^2 - x^2} \left[\frac{x}{\sqrt{R^2 - x^2}}\right]^{1/2} dx$

Then it's somehow simplified to get,

$A = 2\pi \int^R_{-R} \sqrt{R^2 - x^2} \left[\frac{R^2}{\sqrt{R^2 - x^2}}\right]^{1/2}$

Then so on until the final answer is $4 \pi R^2$

The question was to find the surface area of a sphere or radius R. I got a bit stuck so I looked at the solution, however I'm confused as to how the simplification in it was attained. The solution is as follows:

$y = \sqrt{R^2 - x^2}$ in the range $-R \leq x \leq R$

$y' = - \frac{x}{\sqrt{R^2 - x^2}}$

Therefore,

$A = 2\pi \int^R_{-R} \sqrt{R^2 - x^2} \left[\frac{x}{\sqrt{R^2 - x^2}}\right]^{1/2} dx$

Then it's somehow simplified to get,

$A = 2\pi \int^R_{-R} \sqrt{R^2 - x^2} \left[\frac{R^2}{\sqrt{R^2 - x^2}}\right]^{1/2}$

Then so on until the final answer is $4 \pi R^2$

But I have no idea how $\frac{x}{\sqrt{R^2 - x^2}}$ became $\frac{R^2}{\sqrt{R^2 - x^2}}$

Any help please?

That first equation is incorrect.

joostan is almost correct. The formula for the surface area (when rotated about the x-axis) is:


$= 2\pi \displaystyle\int^R_{-R}y \sqrt{1+ \left(\dfrac{dy}{dx} \right)^2} \ dx$

And subbing in your values gives:

$A= 2\pi \displaystyle\int^R_{-R}\sqrt{R^2-x^2} \sqrt{1+ \dfrac{x^2}{R^2-x^2}} \ dx$

Then put what's under the second square root over a common denominator.

Yeah when the function is rotated around the x axis i.e y isn't constant is what I mean.

Hello,

S=2pi*R*integrale(-R;R)dz

but dtheta=2pi*root(R^2-z^2)*dz/cos(theta)=2pi*root(R^2-z^2)*(R*dz)/root( R^2-z^2)=2*pi*R*dz

so S=2*pi*R^2