Gravity due a hemispherical planet

[Yalanhar]

Homework Statement

find the acceleration due to gravity at the centre of a solid hemisphere.

Relevant Equations

$dg=\frac {GMdm}{r^2}$

Can someone help me? I am not sure where is my mistake

correct answer:
$g=\pi G\rho R$

my calculus
By symmetry, I only need to add the vertical component of gravity
$dg = \frac {GMdm}{d^2}\cdot cos\theta$ (1) Where d=R
$\rho =\frac{dm}{dV}$
$dm = \rho \cdot dV$ (2)

(2) in (1)

$dg = \frac {G \rho cos\theta dV}{R^2}$

for spherical coordinate $$dV = r^2drsin\theta d\theta d\phi$$

so:
$dg=\frac {G\rho }{R^2} r^2dr sin\theta cos\theta d\theta d\phi$

$g=\frac {G\rho}{R^2} \int_0^R r^2 \, dr \int_0^\frac {\pi}{2} sin\theta cos\theta \, d\theta \int_0^{2\pi} \, d\phi$

$g = \frac {G\rho}{R^2} \cdot \frac{R^3}{3} \cdot \frac {1}{2} \cdot 2\pi$

$g = \frac {G\rho R\pi}{3}$

$g=\frac {\pi G\rho R}{3}$

 

[kuruman]

Can someone help me? I am not sure where is my mistake

Here is your mistake.

$dg = \frac {G \rho cos\theta dV}{R^2}$

Not all elements $dV$ are at distance $R$ from the origin. The denominator should be $r^2$.

 

[Yalanhar]

Here is your mistake.

Not all elements $dV$ are at distance $R$ from the origin. The denominator should be $r^2$.

oh yes. I got it now. Thanks!