I am trying to solve this question:
'Obtain the surface area of part of the sphere $x^2+y^2+z^2=a^2$ contained within the cone $z \tan \alpha= \sqrt{x^2+y^2}$ where $0 \le \alpha \le \pi/2$.
Explain how you can get the area of the hemisphere using this result.'
I obtained the normal to be $(x,y,z)/a$ and tried to integrate: integration of $$ ds= \int\!\!\!\int a/z \,dx \,dy $$. However I do not know how to obtain the limits. Does anyone know how to solve the problem please?
You can recast into spherical polar coordinates. Since the cone is aligned with the $z$ axis (we can always rotate so it is) we know it will intercept the sphere on a circle of constant $z$. We know that on this circle the radius $a$ forms the hypotenuse and $z$ the adjacent of a right angle triangle, as shown in the diagram below:
The angle that the cone makes with the $x$ axis is precisely $\alpha$ because the slope of this line is $\tan(\alpha)$. Therefore the adjacent angle is given by $\Theta = \frac{\pi}{2}-\arctan(\tan(\alpha)) = \frac{\pi}{2} - \alpha$,
Now we can form the integral in spherical polar coordinates with nice easy limits
$$\int_0^{2\pi} \int_0^\Theta a^2 \sin(\theta)\, d\theta\, d\phi$$
which after evaluation and minor simplification gives us that the surface area is
$$2\pi\, a^2(1-\sin(\alpha))$$
Another way to arrive at this answer is to consider the surface area of the whole sphere multiplied by the ratio of the solid angle formed by the cone to the whole sphere.