Integration problem to calculate surface area of a sphere by rotating the half-circle. Equation to integrate and entire problem statement in next box.

Question

Integration problem to calculate surface area of a sphere by rotating the half-circle.  Equation to integrate and entire problem statement in next box.

anonymous · Answer

$$S = 2 \pi \int\limits   \sqrt{R^2-x^2} \  \sqrt{1 + \frac{x^2}{1 - x^2}} \ dx$$

anonymous · Answer

R is a constant.

amistre64 · Answer

$$\int \sqrt{\frac{R^2-x^2}{1-x^2}}dx$$is what i get to by simplifying ... not that it helps :)

anonymous · Answer

Right, I forget about that step!!

anonymous · Answer

forgot

amistre64 · Answer

$$y' = \frac{R-x}{y}$$
$$(y')^2 = \frac{(R-x)^2}{y2}=\left( \frac{R-x}{R^2-x^2}\right)^2$$trying to see if this is a "given" as stated

amistre64 · Answer

ack, i treated R as a variable lol

amistre64 · Answer

still y^2 = R^2-x^2; not 1-x^2

anonymous · Answer

Thanks for any help.  I'll think about it today.  Funny part is I know the answer has to be Rx.

amistre64 · Answer

you used x^2+y^2 = 1  in the second general part there; when the Radius is spose to be R

amistre64 · Answer

$$\int \sqrt{\frac{R^2-x^2}{(\cancel{1})R^2-x^2}}dx$$

anonymous · Answer

Not following.  The surface element does not contain R, only the first term which is f(x) and that has R^2.

anonymous · Answer

OHHH

anonymous · Answer

I subbed for y in ds

amistre64 · Answer

the circle is:  x^2 + y^2 = R^2
                      y = sqrt(R^2 - x^2)

since y' = -x/y,  and (y')^2 = .....

anonymous · Answer

Thank you!

amistre64 · Answer

youre welcome

amistre64 · Answer

$$\int \sqrt{\frac{R^2(R^2-x^2)}{R^2-x^2}}dx=2piR\int_{-R}^{R} dx$$