Multivariable calculus

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

Q: Why is the value so small?

Two small numbers multiply. The region has area \int_0^1\left(\sqrt{y} - y^2\right)\,dy = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}, and the integrand y^2 never exceeds 1 on it, averaging \frac{3}{35} \cdot 3 = \frac{9}{35} \approx 0.257. One third of an area times a quarter of a unit height is under a tenth.

Q: Do I need to split the region anywhere?

No. The curves cross only at y = 0 and y = 1, and between those values their left-right order never changes, so one double integral covers the whole region.

The double integral of y^2 over the region between the curves x = y^2 and x = sqrt(y), for y from 0 to 1, equals 3/35, about 0.085714. For 0 < y < 1 the curve x = sqrt(y) lies to the right of x = y^2, so it supplies the upper limit of the inner x integral.

f(x,y) = y^2

\int_0^1 \int_{y^2}^{\sqrt{y}} y^2\,dx\,dy = \frac{3}{35}

Numerically 0.085714, confirmed by quadrature on every build.

Horizontal slices, so x runs between two curves

Both boundaries are handed to you as $x$ in terms of $y$ , which is the signal for a type 2 setup: $y$ runs between numbers, and for each $y$ the slice runs left to right between two curves. They meet where $y^2 = \sqrt{y}$ , that is $y^4 = y$ , so $y = 0$ and $y = 1$ .

Test an interior value. At $y = \frac14$ the left curve gives $x = 0.0625$ and the right gives $x = 0.5$ , so $x = \sqrt{y}$ is the right-hand boundary.

\int_{y^2}^{\sqrt{y}} y^2\,dx = y^2\left[x\right]_{x=y^2}^{x=\sqrt{y}} = y^{5/2} - y^4

The integrand carries no $x$ , so the inner step is height times width. The outer integral is then a pair of power rules with a fractional exponent.

\int_0^1 \left(y^{5/2} - y^4\right)\,dy = \frac{2}{7} - \frac{1}{5} = \frac{3}{35}

The mistake: sketching x = y^2 as an upward parabola

$x = y^2$ opens to the right, not upward. Sketch it as the upward parabola $y = x^2$ and you have drawn the other boundary of this region, since in the first quadrant $y = x^2$ and $x = \sqrt{y}$ are the same curve. The two boundaries then trade places in your picture, the inner limits come out as $\int_{\sqrt{y}}^{y^2}$ , and the answer arrives as $-\frac{3}{35}$ .

The reliable test is arithmetic rather than drawing. Pick any $y$ strictly between $0$ and $1$ and compare the two $x$ values. Since $0 < y < 1$ forces $y^2 < y < \sqrt{y}$ , the square is always on the left.

$\sqrt{y} = y^{1/2}$ , so $y^2 \cdot y^{1/2} = y^{5/2}$ , and $\int y^{5/2}\,dy = \frac{2}{7}y^{7/2}$ .
Fractional exponents add like any others: $\frac12 + 2 = \frac52$ , then integrating raises it to $\frac72$ .
The coefficient $\frac{2}{7}$ is $\frac{1}{7/2}$ , a place where switching to decimals hides errors.

The other trap is the outer limits. They must be the $y$ values where the curves intersect, $0$ and $1$ , not $x$ values read off the picture.

The region is symmetric, so type 1 works too

Reflecting in the line $y = x$ maps this region onto itself, so it is type 1 as well: $0 \le x \le 1$ with $y$ running from $x^2$ up to $\sqrt{x}$ . The integrand $y^2$ is not symmetric, so the arithmetic differs, but the answer must not.

\int_0^1 \int_{x^2}^{\sqrt{x}} y^2\,dy\,dx = \int_0^1 \frac{x^{3/2} - x^6}{3}\,dx = \frac{1}{3}\left(\frac{2}{5} - \frac{1}{7}\right) = \frac{3}{35}

Two different routes to $\frac{3}{35}$ is a strong check. This one is slightly heavier because cubing the fractional bound produces $x^{3/2}$ .

Frequently asked questions

Why is the value so small?

Two small numbers multiply. The region has area $\int_0^1\left(\sqrt{y} - y^2\right)\,dy = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ , and the integrand $y^2$ never exceeds $1$ on it, averaging $\frac{3}{35} \cdot 3 = \frac{9}{35} \approx 0.257$ . One third of an area times a quarter of a unit height is under a tenth.

Do I need to split the region anywhere?

No. The curves cross only at $y = 0$ and $y = 1$ , and between those values their left-right order never changes, so one double integral covers the whole region.