Show the triple integral given is equivalent to $\frac{15\pi}{16}$ Evaluate $$\iiint_E\;z \, dV$$
where E is enclosed between the spheres $x^2 + y^2 + z^2 = 1$and$x^2 + y^2 + z^2 = 4$ in the first octant.
I'll be honest. My first attempt didn't worked very well. My answer showed for my first attempt was 7pi/12 because the integral I set up was like this:
$$\iiint _E\;z\;dV\\
1\leq \rho\leq 2\\
0\leq\theta\leq \frac{1}{2}\pi\\
0\leq\phi\leq \frac{1}{2}\pi$$
Which transforms the integral into
$$\int_{0}^{\frac{1}{2}\pi}\int_{0}^{\frac{1}{2}\pi}\int_{1}^{2}(\rho)^2 cos\phi\sin\phi\ d\rho\; d\theta \;d\phi\\$$
This problem was taken from Stewart Calculus 6th Edition. Like I said, I tried to solve this problem on my own and got an answer of $$\frac {7}{12}\pi$$
but when I browsed the answers section, it says its answer was $$\frac {15}{16}\pi$$
Did I do something wrong in my computation?

Hint. Observe that the converted integral is rather
$$
\int_{0}^{\pi/2}\!\int_{0}^{\pi/2}\!\int_{1}^{2}\rho\cos\phi \:\rho^2\sin\phi\ d\rho\; d\theta \;d\phi=\int_{1}^{2}\rho^3d\rho\times \int_{0}^{\pi/2}d\theta\times \int_{0}^{\pi/2}\frac{\sin(2\phi)}2 d\phi=\frac{15}{16}\pi
$$ as announced.
