Find the volume of a torus. (The solid generated

Find the volume of a torus. (The solid generated by rotating the circle $x^2 + (y - b)^2 = a^2$ about the $x$-axis, where $b > a$.)

We compute the torus's volume with the washer method, then confirm it instantly via Pappus's theorem.

Step 1: Geometry

The circle $x^2+(y-b)^2=a^2$ is centered at $(0,b)$ with radius $a$; since $b>a$ it never touches the $x$-axis, so revolving it about the $x$-axis produces a torus (donut shape) rather than a self-intersecting solid.

Step 2: Boundary functions

\[y = b\pm\sqrt{a^2-x^2}, \qquad -a\le x\le a\]

Upper boundary $y_+=b+\sqrt{a^2-x^2}$; lower boundary $y_-=b-\sqrt{a^2-x^2}$.

Step 3: Washer method

At position $x$, the washer has outer radius $R(x)=y_+$ and inner radius $r(x)=y_-$:

\[A(x) = \pi\left(y_+^2-y_-^2\right) = \pi(y_++y_-)(y_+-y_-)\]

Writing $s=\sqrt{a^2-x^2}$: $y_++y_-=2b$ and $y_+-y_-=2s$, so:

\[A(x) = \pi\cdot2b\cdot2s = 4\pi b\sqrt{a^2-x^2}\]

Step 4: Integrate

\[V = 4\pi b\int_{-a}^{a}\sqrt{a^2-x^2}\,dx\]

The remaining integral is the area under the upper half of the circle $x^2+v^2=a^2$, i.e. the area of a semicircle of radius $a$:

\[\int_{-a}^{a}\sqrt{a^2-x^2}\,dx = \frac{\pi a^2}{2}\]

Step 5: Calculate the volume

\[V = 4\pi b\cdot\frac{\pi a^2}{2} = 2\pi^2 a^2b\]

\[\boxed{V = 2\pi^2 a^2b \text{ cubic units}}\]

Step 6: Cross-check via Pappus's centroid theorem

Pappus's theorem states that the volume swept by a plane region revolved about an external axis equals the region's area times the distance traveled by its centroid. Here the region is the disk of radius $a$ (area $\pi a^2$) with centroid at $(0,b)$, which travels a circular path of length $2\pi b$:

\[V = \pi a^2\times2\pi b = 2\pi^2a^2b \quad\checkmark\]

Summary

\[\boxed{V = 2\pi^2 a^2b \text{ cubic units}}\]

Key insight: The torus volume factors neatly as $(\pi a^2)\times(2\pi b)$ — cross-sectional area times the circumference traced by the tube's center — a direct and elegant illustration of Pappus's theorem.