Multivariable calculus

Triple Integral of x/(1 + y^2) over a Box

Q: Why does the missing z variable still change the answer?

Integrating a z free function over the z edge multiplies by the length of that edge, here 3. Only an edge of length 1 leaves the value untouched, and an edge of length 0 would collapse the solid and the integral with it.

Q: Is the integrand ever undefined on this box?

No. The denominator 1 + y^2 is at least 1 for every real y, so the integrand is smooth and bounded everywhere on the box. That is what makes this a safe example, unlike \frac{x}{y^2 - 1}, which blows up at y = 1.

The triple integral of x divided by 1 + y squared, over the box where x runs 0 to 2, y runs 0 to 1 and z runs 0 to 3, equals 3 pi over 2, about 4.712389. The x factor gives 2, the y factor is arctan(1) which is pi/4, and the z factor gives 3.

f(x,y,z) = \frac{x}{1 + y^2}

\int_0^2 \int_0^1 \int_0^3 \frac{x}{1 + y^2} \,dz\,dy\,dx = \frac{3\pi}{2}

Numerically 4.712389, confirmed by quadrature on every build.

The variable that is missing does the least work

No $z$ appears in the integrand, so the inner integral in $z$ just multiplies by the length of that edge. That is the fastest stage on the page and it is easy to skip by accident.

\int_0^3 \frac{x}{1 + y^2} \,dz = \frac{3x}{1 + y^2}

What is left separates. The $x$ factor is a plain power rule and the $y$ factor is the standard arctangent form.

\left(\int_0^2 x \,dx\right)\left(\int_0^1 \frac{dy}{1 + y^2}\right)(3) = 2\cdot\arctan(1)\cdot 3 = 2\cdot\frac{\pi}{4}\cdot 3

That product is $\frac{6\pi}{4} = \frac{3\pi}{2} \approx 4.712389$ .

The mistake: turning the denominator into a logarithm

Seeing $\frac{1}{1 + y^2}$ triggers the reflex $\ln\left(1 + y^2\right)$ , but that antiderivative belongs to $\frac{2y}{1 + y^2}$ , where the numerator is the derivative of the denominator. Here the numerator is $1$ , so no substitution applies and the answer is $\arctan y$ .

$\int \frac{dy}{1 + y^2} = \arctan y + C$ , giving $\frac{\pi}{4}$ over $[0, 1]$ .
$\int \frac{2y \,dy}{1 + y^2} = \ln\left(1 + y^2\right) + C$ , giving $\ln 2 \approx 0.693$ over $[0, 1]$ .
The gap is $\arctan(1) = 0.785398$ against $\ln 2 = 0.693147$ , close enough to look plausible and wrong enough to lose the question.

Differentiating your candidate answer takes five seconds and separates the two immediately.

Why a pi shows up with no circle in sight

The region is a rectangular box and the integrand is a rational function, so a $\pi$ in the answer surprises people. It arrives through $\arctan$ , whose values at nice inputs are angles, and $\arctan(1) = \frac{\pi}{4}$ is the angle of a $45$ degree line.

Change the $y$ edge and the $\pi$ can disappear from the neat form. Running $y$ from $0$ to $\tan(1)$ would give $\arctan(\tan 1) = 1$ instead, and the answer becomes exactly $6$ .

Frequently asked questions

Why does the missing z variable still change the answer?

Integrating a $z$ free function over the $z$ edge multiplies by the length of that edge, here $3$ . Only an edge of length $1$ leaves the value untouched, and an edge of length $0$ would collapse the solid and the integral with it.

Is the integrand ever undefined on this box?

No. The denominator $1 + y^2$ is at least $1$ for every real $y$ , so the integrand is smooth and bounded everywhere on the box. That is what makes this a safe example, unlike $\frac{x}{y^2 - 1}$ , which blows up at $y = 1$ .