This is wrong but I can't figure out what is wrong, could someone help me find the errors.
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\$\begingroup\$ The electric field at distance X is just an inverse function of \$X^2\$ isn't it? hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html \$\endgroup\$ – Andy aka Feb 11 '17 at 20:03
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2\$\begingroup\$ It would help a lot if you could write the question. Rho is given as surface charge, but you treat it as volume charge. What is it? \$\endgroup\$ – sweber Feb 11 '17 at 23:40
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\$\begingroup\$ \$\bar r^{'}\$ does not always line up with \$\bar r\$, so you cannot assign the same \$\bar a_r\$ to both of them. \$\endgroup\$ – rioraxe Feb 12 '17 at 0:22
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What's missing here is the original question. But I guess it is something like
For a hollow sphere with radius \$r_0\$ and surface charge \$\rho_s=1000C/m^2\$ located at the origin of the coordinate system, find the electric field at a point \$\vec{x}_0\$ outside the sphere.
In general, you have two issues in your calculation.
First, it is a surface charge, and \$dq\$ does not depend on any \$dr\$, just \$r^2_0\$:
$$ds=r_0^2\sin\theta\,d\theta\,d\phi$$
$$dq=\rho_sr_0^2\sin\theta\,d\theta\,d\phi$$
So you'll only have to integrate over this two angles, too:
$$\iint...r_0^2\sin\theta\,d\theta\,d\phi$$
Second each small charge \$dq\$ located at a position \$dr\$ creates a tiny field at \$\vec{x}_0\$. Direction and strength depend on the distance between both:
$$\vec{R}=\vec{x}_0-\vec{r}$$
Of course, \$r\$ is different for each \$dq\$, and since you have chosen spherical coordinates, you have to express this in spherical coordinates, too:
$$\vec{r}=\begin{pmatrix}r\,\sin \theta \,\cos \varphi \\r\,\sin \theta \,\sin \varphi \\r\,\cos \theta \end{pmatrix}$$
With this, the integral is
$$\vec{E}(\vec{x}_0)=\iint\frac{1}{4\pi\varepsilon_0}\frac{\vec{R}}{|\vec{R}|^3}\,dq$$
$$=\iint\frac{1}{4\pi\varepsilon_0}\frac{\vec{R}}{|\vec{R}|^3}\cdot\rho_sr_0^2\sin\theta\,d\theta\,d\phi=\frac{\rho_sr_0^2}{4\pi\varepsilon_0}\iint\frac{\vec{R}}{|\vec{R}|^3}\sin\theta\,d\theta\,d\phi$$
Finally, should you really calculate the field this way? The integral is a bit nasty... Typically, one uses Gauß law plus geometry to get the well known result
$$\vec{E}(\vec{x}_0)=\frac{q}{4\pi\varepsilon_0}\frac{\vec{x}_0}{|\vec{x}_0|^3}$$
with
$$q=4\pi r_0^2\rho_s$$
EDIT: Lost the \$\sin\theta\$...
