The solid shown below has a semicircular base of 2cm. Vertical cross-sections of the solid perpendicular to the diameter of the semicircle are right-angled triangles, the heights of which are bounded by the parabola $y=4-x^{2}$. Show that the solid has volume $3\pi \space cm^{3}$.
I understand the height of the right-angled triangle is given by the y-coordinate of the parabola and the base of the triangle is given by the y-coordinate of the semicircle. So, technically, the area of the triangle should be $\frac{1}{2}y_{parabola}y_{semicircle}=\frac{1}{2}(4-x^{2})\sqrt{4-x^2}$. Therefore, $\delta V=\frac{1}{2}(4-x^{2})\sqrt{4-x^2}\delta x$. Following this, I would go on to do the following: $\lim_{\delta x\to0}\sum_{-2}^{2}(\frac{1}{2}(4-x^{2})\sqrt{4-x^2}\delta x)=\int_{-2}^{2}\frac{1}{2}(4-x^{2})\sqrt{4-x^2}dx$. However, to do this, I am assuming there are two y-axes, which I do not think is possible. So how am I required to go about solving this?
The other axis (with the semicircle) is not labeled on your graph but I agree that it is a bit strange and confusing to call it (a) $y$-axis if that's already used for the vertical one (with the parabola).
However, no matter how you call it, e.g. a $z$-axis, the base of the triangle as a function of $x$ is still $\sqrt{4-x^2}$, e.g. $z=\sqrt{4-x^2}$, so the rest of the calculation remains the same - and looks correct to me!
$$\frac{1}{2}\int_{-2}^{2}\underbrace{\left(4-x^{2}\right)}_{\mbox{height $y(x)$ }}\;\underbrace{\sqrt{4-x^2}}_{\mbox{base $z(x)$ }}\,\mbox{d}x = \ldots = \boxed{3\pi}$$