Multivariable calculus

Double Integral of y Between y = x^2 and y = x

Q: Does the answer change in the order dx dy?

No, it is still \frac{1}{15}, but the setup flips. For y in [0,1] the slice runs from x = y on the left to x = \sqrt{y} on the right, since on (0,1) the parabola lies to the right of the line. The integral becomes \int_0^1 y\left(\sqrt{y} - y\right)\,dy = \frac{2}{5} - \frac{1}{3} = \frac{1}{15}.

Q: Why integrate the bare function y?

It is the integrand that produces the moment about the x-axis, and it also keeps the inner step transparent: the result is \frac{\text{top}^2 - \text{bottom}^2}{2}, which is a statement about the region rather than about the integrand.

The double integral of y over the region between the parabola y = x^2 and the line y = x, for x from 0 to 1, equals 1/15. On that interval the line is above the parabola, so the inner limits run from y = x^2 up to y = x.

f(x,y) = y

\int_0^1 \int_{x^2}^{x} y\,dy\,dx = \frac{1}{15}

Numerically 0.066667, confirmed by quadrature on every build.

Find where the curves meet, then slice

Set $x^2 = x$ to get $x = 0$ and $x = 1$ . Those intersections are the outer limits. Between them, test one interior point: at $x = \frac12$ the line gives $0.5$ and the parabola gives $0.25$ , so the line is the top boundary.

\int_{x^2}^{x} y\,dy = \left[\frac{y^2}{2}\right]_{y=x^2}^{y=x} = \frac{x^2 - x^4}{2}

\int_0^1 \frac{x^2 - x^4}{2}\,dx = \frac{1}{2}\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{1}{2}\cdot\frac{2}{15} = \frac{1}{15}

With $f = y$ the inner step is pure geometry: it returns half the difference of the squared boundary heights, so any error in identifying top and bottom shows up immediately.

The mistake: putting the parabola on top

Outside $[0,1]$ the parabola is the higher curve, so the ordering people carry over from other problems is backwards here. Swapping the limits negates the answer to $-\frac{1}{15}$ .

That negative is a free error detector. The integrand $y$ is non-negative everywhere on this region, so a negative result is a contradiction you can spot without rechecking a single line of algebra.

At $x = \frac12$ : line $0.5$ , parabola $0.25$ . The line wins on $(0,1)$ .
At $x = 2$ : line $2$ , parabola $4$ . The parabola wins outside.
The curves cross only at $x = 0$ and $x = 1$ , so no splitting is needed in between.

A related slip is integrating $y$ with respect to $x$ by reflex, producing $xy$ and leaving a $y$ stranded in the outer integrand. The differential names the variable, not the letters in the integrand.

What 1/15 measures

This integral is the first moment of the region about the $x$ -axis. Divide by the area and you get the height of the centroid.

\text{area} = \int_0^1 \left(x - x^2\right)\,dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}, \qquad \bar{y} = \frac{1/15}{1/6} = \frac{2}{5}

A centroid height of $0.4$ is believable. The region is a thin lens between the line and the parabola: at $x = \frac12$ it spans $0.25 \le y \le 0.5$ , and it only reaches heights near $1$ as $x$ approaches $1$ .

Frequently asked questions

Does the answer change in the order dx dy?

No, it is still $\frac{1}{15}$ , but the setup flips. For $y$ in $[0,1]$ the slice runs from $x = y$ on the left to $x = \sqrt{y}$ on the right, since on $(0,1)$ the parabola lies to the right of the line. The integral becomes $\int_0^1 y\left(\sqrt{y} - y\right)\,dy = \frac{2}{5} - \frac{1}{3} = \frac{1}{15}$ .

Why integrate the bare function y?

It is the integrand that produces the moment about the $x$ -axis, and it also keeps the inner step transparent: the result is $\frac{\text{top}^2 - \text{bottom}^2}{2}$ , which is a statement about the region rather than about the integrand.