I have to prove the surface area of a sphere with $r=1$ using the solids of revolution through revolution abouth both the $x$ and the $y$ axis.
The formulas are easy. From top to bottom, surface area of revolution about $x$ axis, and $y$ axis formulas:
$$S_x=\int_a^b2\pi y\,\sqrt{1+\Big(\frac{dy}{dx}\Big)^2}\,dx$$ $$S_y=\int_a^b2\pi x\,\sqrt{1+\Big(\frac{dx}{dy}\Big)^2}\,dy$$ Where in the first formula, $y$ is the function of $x$, in the second one, $x$ is the function of $y$, and in both $a$ and $b$ is the section of the function to rotate.
With these formulas, I need to prove, as I stated, the surface area of a sphere with radius $1$ by rotating about both axes. But to do that, I need to have some function that has a perfect semi-circle in it that is two units long/high, so it has diameter of $2$/radius of $1$.
I just don't know what function has a perfect semi-circle in it. Perhaps is it a conic section, that is cut in half of some sort, or I don't know. Does anyone know, what functions could I use that have these properties, so I can prove the surface area the way I was told to?
On a sphere, $x^2 + y^2 + z^2 = 1$, so
$$ z = \sqrt{1 - (x^2 + y^2)} $$ gives an equation for points in the upper hemisphere; $$ z = -\sqrt{1 - (x^2 + y^2)} $$ gives points in the lower hemisphere.
But for the thing you want to rotate, you just need a plot in the xy-plane, and for that,
$$ y = \pm \sqrt{1 - x^2} $$ or $$ x = \pm \sqrt{1 - y^2} $$ will do the trick, as will, using yet another formulat, but one you probably haven't seen,
$$ x(t) = \cos t\\ y(t) = \sin t. $$