15. A square with side a (Fig. 98) rotates around an axis passing through its vertex and forming an angle $\varphi\left(0 \leqslant \varphi \leqslant 45^{\circ}\right)$ with its side.

Find the volume and surface area of the resulting solid of revolution. (O. 117, P. 210.)
15. Let $O$ be the center of the square (see Fig. 98), $R$ be the distance from the axis of rotation to $O$. Then

$$
R=A O \cdot \sin \left(\varphi+45^{\circ}\right)=\frac{a \sqrt{2}}{2} \sin \left(\varphi+45^{\circ}\right)
$$

The volume $V$ and surface area $S$ of the solid of revolution are found using Guldin's theorems:

$$
\begin{gathered}
V=2 \pi R \cdot a^{2}=2 \pi \frac{a \sqrt{2}}{2} \sin \left(\varphi+45^{\circ}\right) a^{2}= \\
=\pi a^{3} \sqrt{2} \cdot \sin \left(\varphi+45^{\circ}\right) \\
S=2 \pi R \cdot 4 a=4 \pi a^{2} \sqrt{2} \sin \left(\varphi+45^{\circ}\right)
\end{gathered}
$$