Answer
$${2 \pi^{2} r^{2} h}$$
Work Step by Step
The volume of the torus is the volume obtained by the washer perpendicular to the $y$-axis, as the circle equation shown in the fig is $r^{2}=(x-h)^{2}+y^{2}$ The equation of the right half of the circle is $h+\left(r^{2}-y^{2}\right)^{1 / 2}=x$ , the equation of the left half of the circle $h-\left(r^{2}-y^{2}\right)^{1 / 2}=x$ The revolving washer is enclosed between these two curves Hence $h+\left(r^{2}+y^{2}\right)^{1 / 2}=w(y)$ and $h-\left(r^{2}-y^{2}\right)^{1 / 2}=v(y)$
Thus, the volume of the torus obtained by rotating this disc perpendicular to $y$-axis is as follows.
$$
\begin{aligned}
\int_{-r}^{r} \pi\left[[w(y)]^{2}-[v(y)]^{2}\right] d y=V & \\
&=\pi \int_{-r}^{r}\left[h+\left(r^{2}+y^{2}\right)^{1 / 2}\right]^{2}-\left[h-\left(r^{2}-y^{2}\right)^{1 / 2}\right]^{2} \\
&=\pi \int_{-r}^{r} 4 h\left(r^{2}-y^{2}\right)^{1 / 2} d y \\
&=4 h \pi \int_{-r}^{r}\left(r^{2}-y^{2}\right)^{1 / 2} d y
\end{aligned}
$$
Let $r \sin t=y$ and then $r \cos t d t=d y$ ,$-\pi / 2 \leq t \leq \pi / 2$ we get
$$
\begin{aligned}
4 h \pi \int_{-\pi / 2}^{\pi / 2}\left(r^{2}-r^{2} \sin ^{2} t\right)^{1 / 2}(r \cos t) d t=V & \\
&=4 h \pi \int_{-\pi / 2}^{\pi / 2}\left(r \cos ^{2} t\right) d t \\
&=4 h \pi\left[\frac{1}{2} t+\frac{1}{4} \sin 2 t\right]_{-\pi / 2}^{\pi / 2} \\
&=4 r^{2} h \pi(\pi / 2) \\
&={2 \pi^{2} r^{2} h}
\end{aligned}
$$
