I tried using trig substitution to evaluate this, but I just can seem to arrive at the expected answer of $$V=\frac{2}{3}r^3$$
Because I may be committing mistakes while evaluating, though with the limited knowledge I have I'm not sure what I'm doing wrong.
The limits are: $$-\sqrt{r^2-y^2}\leq x \leq\sqrt{r^2-y^2}$$ $$0\leq z \leq y$$ $$0\leq y \leq r$$
The area of the triangular shape is $$A=\frac{1}{2}(r^2-t^2)$$
Would I be able to use the area to do this quicker or should I just evaluate the limits?
From the previous part of the problem, you have $x=t$ and $-r\le t\le r$, so $-r\le x\le r$, not the range that you have here. Just integrate the cross-section area over this interval: $$\int_{-r}^r \frac12(r^2-x^2)\,dx = \int_0^r r^2-x^2\,dx = \frac23r^3.$$