Multivariable calculus

Double Integral of x/(1+y^2) Over a Triangle

Q: How do I know the hypotenuse is y = x/2?

Two vertices determine it. Through (0,0) and (2,1) the slope is \frac{1-0}{2-0} = \frac12, and the line passes through the origin, so y = \frac{x}{2}. The third vertex (2,0) lies below it, confirming the region sits between y = 0 and that line.

Q: Why does pi appear in the answer at all?

It comes from the arctangent. The upper limit x = 2 makes \arctan(1) = \frac{\pi}{4}, and that value enters twice in the final assembly, once from the boundary term of the parts formula and once from the leftover integral.

The double integral of x/(1 + y^2) over the triangle with vertices (0,0), (2,0) and (2,1) equals pi - 2, about 1.141593. Integrating in y first gives x arctan(x/2), which then needs integration by parts. Integrating in x first is much shorter.

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

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

Numerically 1.141593, confirmed by quadrature on every build.

The inner integral is an arctangent

The slanted edge joins $(0,0)$ to $(2,1)$ , so it is the line $y = \frac{x}{2}$ . For fixed $x$ in $[0,2]$ the vertical slice runs from $y = 0$ to $y = \frac{x}{2}$ , and $x$ is a constant throughout that step.

\int_0^{x/2} \frac{x}{1+y^2}\,dy = x\left[\arctan y\right]_{y=0}^{y=x/2} = x\arctan\!\left(\frac{x}{2}\right)

The outer integral needs parts, with $u = \arctan\!\left(\frac{x}{2}\right)$ and $dv = x\,dx$ . The derivative you need is $\frac{d}{dx}\arctan\!\left(\frac{x}{2}\right) = \frac{2}{4+x^2}$ .

\int_0^2 x\arctan\!\left(\frac{x}{2}\right)\,dx = \left[\frac{x^2}{2}\arctan\!\left(\frac{x}{2}\right)\right]_0^2 - \int_0^2 \frac{x^2}{4+x^2}\,dx

= \frac{\pi}{2} - \int_0^2\left(1 - \frac{4}{4+x^2}\right)\,dx = \frac{\pi}{2} - 2 + \frac{\pi}{2} = \pi - 2

The mistake: reaching for a logarithm

During the inner step the whole factor $x$ is a constant and comes straight out. The wrong move is to see a quotient, assume a logarithm, and write $\ln\!\left(1+y^2\right)$ .

These are two different integrals, and the difference is what sits in the numerator.

Numerator constant in $y$ : $\int \frac{dy}{1+y^2} = \arctan y$ .
Numerator proportional to $y$ : $\int \frac{y\,dy}{1+y^2} = \frac{1}{2}\ln\!\left(1+y^2\right)$ .
Both present: split the fraction into two integrals first.

The second common error is the upper limit. The hypotenuse joins $(0,0)$ to $(2,1)$ , so it is $y = \frac{x}{2}$ , not $y = 2x$ . Using the wrong line turns $\arctan\!\left(\frac{x}{2}\right)$ into $\arctan(2x)$ and changes the answer completely.

Integrating in x first is easier

As a type 2 region the same triangle is $0 \le y \le 1$ with $x$ running from the hypotenuse $x = 2y$ across to $x = 2$ . Now the $x$ integral is a power rule and the leftover $y$ integral is standard.

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

No integration by parts at all. When a region is both type 1 and type 2, spend ten seconds choosing the order before you spend ten minutes integrating.

Frequently asked questions

How do I know the hypotenuse is y = x/2?

Two vertices determine it. Through $(0,0)$ and $(2,1)$ the slope is $\frac{1-0}{2-0} = \frac12$ , and the line passes through the origin, so $y = \frac{x}{2}$ . The third vertex $(2,0)$ lies below it, confirming the region sits between $y = 0$ and that line.

Why does pi appear in the answer at all?

It comes from the arctangent. The upper limit $x = 2$ makes $\arctan(1) = \frac{\pi}{4}$ , and that value enters twice in the final assembly, once from the boundary term of the parts formula and once from the leftover integral.