Double integration calculation

[Fys]

I want to calculate the integral of a sphere with radius a using double integrals

I have $$z^2+y^2+x^2=a$$ and I plug this in my equation and integrate in polar coördinates

Now i have $$\int^{2\pi}_{0} \int ^{a}_{0} \sqrt{(a-r^{2})} rdrd\theta$$

But why do I need to multiply my integral with 2 (because i get half the answer I expect).
Can someone explain this please?

thanks guys

 

[Almanzo]

I assume that it is the volume of the sphere which you wish to calculate.

In polar coordinates, the volume element is dr * (r * dtheta) * {r * sin(theta) * dphi}.
Or r² *dr * sin(theta) * dtheta * dphi.

This must be integrated for r between 0 (the midpoint) and R (the surface), for theta between 0 (the north pole) and pi (the south pole) and phi between zero (setting out from some meridian eastward) and 2*pi (arriving at the same meridian from the west).

The integrals can be split.
Integrating r²dr between 0 and R, we get (1/3)R³.
Integrating sin(theta) dtheta between 0 and pi we get 1 - (-1) = 2.
Integrating dphi between 0 and 2*pi, we get 2*pi.

Multiplying, we get (4/3)*pi*R³.

If you want the surface of the sphere, you use the surface element (R*dtheta) *{R*sin(theta)*dphi}, just do the theta and phi integration, and get 4*pi*R².

Looking at your own method, I get the impression that you are only calculating the volume of the northern hemisphere. The inner integral should run fron -a to +a, not from 0 to +a.