Why cant the height of the element be Rd theta when finding volume since it works when finding surface area.

Step 1
The difference in the method of integration is that in the case of area, you need to measure a slant distance along the surface of the sphere from one circular edge of the area element to the other, whereas in the case of volume you need to measure the perpendicular distance from one circular face of the cylinder to the other.
The differential $R\mathrm{d}\mathrm{\theta <!--\; \theta \; -->}$ accurately represents the slant distance along the surface of the sphere, which is perfect for measuring the surface area, but the surface of the sphere is not perpendicular to the circular faces of the cylindrical element (except when $\mathrm{\theta <!--\; \theta \; -->}=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$), and therefore $R\mathrm{d}\mathrm{\theta <!--\; \theta \; -->}$ is not generally an accurate representation of the height of the cylindrical element.
The plane that cuts the sphere at an angle $\mathrm{\theta <!--\; \theta \; -->}$ from the axis is at a distance $R\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})$ from the center. If you increase the angle to $\mathrm{\theta <!--\; \theta \; -->}+\mathrm{\Delta <!--\; \Delta \; -->}\mathrm{\theta <!--\; \theta \; -->}$ then the distance of the plane from the center is $R\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}+\mathrm{\Delta <!--\; \Delta \; -->}\mathrm{\theta <!--\; \theta \; -->})$. The cylindrical element for the change in angle $\mathrm{\Delta <!--\; \Delta \; -->}\mathrm{\theta <!--\; \theta \; -->}$ is between these two planes, so the height of the cylindrical element is $|<!--\; |\; -->R\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}+\mathrm{\Delta <!--\; \Delta \; -->}\mathrm{\theta <!--\; \theta \; -->})-<!--\; -\; -->R\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})|<!--\; |\; -->$.
Step 2
To use this in an integral, take the limit
$\underset{\mathrm{\theta <!--\; \theta \; -->}\to <!--\; \to \; -->0}{lim}\frac{|<!--\; |\; -->R\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}+\mathrm{\Delta <!--\; \Delta \; -->}\mathrm{\theta <!--\; \theta \; -->})-<!--\; -\; -->R\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->})|<!--\; |\; -->}{\mathrm{\Delta <!--\; \Delta \; -->}\mathrm{\theta <!--\; \theta \; -->}}=R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->},$, which tells you that the perpendicular distance between circular cross-sections is represented by $R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\mathrm{d}\mathrm{\theta <!--\; \theta \; -->}.$.
So the correct integral for the volume is ${\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}}\mathrm{\pi <!--\; \pi \; -->}(R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2(R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\mathrm{d}\mathrm{\theta <!--\; \theta \; -->}).$.
If you still do not see how to correctly translate the angle $\mathrm{\theta <!--\; \theta \; -->}$ into the height of a cylindrical cross-section, try integrating by a variable that translates directly into the height of each cylinder. Put the sphere at the origin of a Cartesian system of coordinates and integrate along the z axis. The radius of the cylindrical element for any given value of z is $\sqrt{R^2-<!--\; -\; -->z^2}$ and the height of the element is dz, so the integral is
${\int <!--\; \int \; -->}_{-<!--\; -\; -->R}^R\sqrt{R^2-<!--\; -\; -->z^2}\mathrm{d}z.$
Make the trig substitution $z=-<!--\; -\; -->R\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$ to get an integral in terms of $\mathrm{\theta <!--\; \theta \; -->}.$. Note that in this substitution, you will replace dz with $R\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\mathrm{d}\mathrm{\theta <!--\; \theta \; -->}.$.