If $\displaystyle u = \cos\phi$ then $\displaystyle u = \cos0 = 1$ when $\displaystyle \phi=0$, and $\displaystyle u = \cos(\pi/4) = 1/\sqrt2$ when $\displaystyle \phi = \pi/4.$ That gives $\displaystyle V = -\frac{2\pi}3\int_1^{1/\sqrt2}u^3\,du.$ From there, the best thing to do is probably to switch the limits of integration, which changes the sign of the integral and leads to $\displaystyle V = \frac{2\pi}3\int^1_{1/\sqrt2}u^3\,du.$ That should be easy enough to integrate.