# The volume of a sphere in spherical coordinates

Using the following formula for a sphere i tried to calculate the volume: $x^2+y^2+z^2=a^2$ and using the fact that $x^2+y^2+z^2 =R^2$ \begin{align*} \int\int\int x^2+y^2+z^2 dxdydz &= \int_0^{2\pi}\int_0^{\frac{1}{2}\pi}\int_0^a R^2 \sin(\phi) R^2 dRd\phi\theta\\ &= \frac{4\pi}{5}a^5 \end{align*} However the actual formula for volume is $\frac{4\pi}{3}a^3$ why does one R^2 disappear? One comes from the jacobian matrix and one from the $x^2+y^2+z^2 =R^2$.

• Integral to find volume does not have $x^2+y^2+z^2$. It's just $\int dV$. – Paul Mar 2 '17 at 20:19
• Ofcourse it does, i could not figure it out for 4 hours. Thanks alot ( i feel silly now) – R.vW Mar 2 '17 at 20:22
• The area of a unit square, for example, is $\int_0^1\int_0^1 dydx$. – Théophile Mar 2 '17 at 20:22