Volume of revolution for a semicircle about the y-axis

Expert intro: This is a classic volume of revolution problem using geometry or the disk method.

Detailed walkthrough

Identify the region

The curve is

$y^2=a^2-x^2,\quad x\ge 0,$

which is the right half of the circle $x^2+y^2=a^2$. The region enclosed by this curve and the $y$-axis is the right semicircular disk of radius $a$.

When that region is rotated about the $y$-axis through $2\pi$ radians, it generates a solid sphere of radius $a$.

Method 1: Use geometry

Because a full rotation of a semicircular region about its diameter produces a sphere, the volume is immediately

$V=\frac{4}{3}\pi a^3.$

This is the fastest route if the geometry is recognized correctly.

Method 2: Use the disk method

Write the curve as

$x=\sqrt{a^2-y^2}.$

At height $y$, the radius of the disk is $x$, so the cross-sectional area is

$\pi x^2=\pi(a^2-y^2).$

The limits for $y$ are from $-a$ to $a$, so

$V=\int_{-a}^{a}\pi(a^2-y^2)\,dy.$

Now integrate:

$V=\pi\left[a^2y-\frac{y^3}{3}\right]_{-a}^{a}$

$=\pi\left(\left(a^3-\frac{a^3}{3}\right)-\left(-a^3+\frac{a^3}{3}\right)\right)$

$=\pi\left(\frac{2a^3}{3}+\frac{2a^3}{3}\right)=\frac{4}{3}\pi a^3.$

Key idea

The important recognition is that the given curve is not just any algebraic relation; it describes a semicircle. Once the region is seen as a half-disk, the rotation about the $y$-axis becomes the standard sphere-volume result.

If you do use integration, remember that the variable $y$ is the natural integration variable because the axis of rotation is vertical and the radius is expressed as a function of $y$.

Final answer

$\boxed{\frac{4}{3}\pi a^3}$

💡 Pitfall guide

A common mistake is to rotate the wrong region mentally and treat the shape as if it were a cylinder or cone. Another error is to set up the integral with the wrong bounds. Since the curve is the right half of a circle, the $y$-values run from $-a$ to $a$, not from $0$ to $a$ unless the problem has explicitly restricted the region. Also, because the rotation is around the $y$-axis, the radius in a disk method setup must be measured horizontally as a function of $y$.

🔄 Real-world variant

If the same region were rotated about the $x$-axis instead, the solid would still have volume $\frac{4}{3}\pi a^3$ because the region is a semicircle of radius $a$ and the axis passes through its center. But if only the upper half of the right semicircle were used, the bounds would change and the solid would no longer be a full sphere. In that case, the region description must be checked carefully before integrating.

🔍 Related terms

volume of revolution, disk method, sphere volume