Multivariable calculus

Double Integral of x^2 y Under y = x^2

Q: Why does a 7 appear in the denominator?

Because the inner integral raises the degree to x^6, and integrating x^6 produces \frac{x^7}{7}. The 7 is a fingerprint of the parabolic top: over a rectangle the same integrand would leave only the powers the integrand itself supplies.

Q: What is the area of this region, and the average value there?

The area is \int_0^2 x^2\,dx = \frac{8}{3}. Dividing the integral by it gives the average of x^2 y on the region, \frac{64}{7} \cdot \frac{3}{8} = \frac{24}{7} \approx 3.43.

The double integral of x^2 y over the region under the parabola y = x^2 with 0 <= x <= 2 equals 64/7, about 9.142857. The inner integral in y gives x^6/2, and integrating that from 0 to 2 gives 128/14, which reduces to 64/7.

f(x,y) = x^2 y

\int_0^2 \int_0^{x^2} x^2 y\,dy\,dx = \frac{64}{7}

Numerically 9.142857, confirmed by quadrature on every build.

Vertical slices under the parabola

Under $y = x^2$ between $x = 0$ and $x = 2$ means each vertical slice starts on the $x$ -axis and stops on the parabola, so the inner limits are $y = 0$ and $y = x^2$ .

\int_0^{x^2} x^2 y\,dy = x^2\left[\frac{y^2}{2}\right]_{y=0}^{y=x^2} = x^2 \cdot \frac{x^4}{2} = \frac{x^6}{2}

Watch the degree climb. Substituting $y = x^2$ into $\frac{y^2}{2}$ produces $\frac{x^4}{2}$ , and the $x^2$ that behaved as a constant during the inner step multiplies it to give $x^6$ .

\int_0^2 \frac{x^6}{2}\,dx = \left[\frac{x^7}{14}\right]_0^2 = \frac{128}{14} = \frac{64}{7}

The mistake: substituting the limit without squaring it

The inner antiderivative is $\frac{y^2}{2}$ and the upper limit is $y = x^2$ , so the substitution gives $\frac{(x^2)^2}{2} = \frac{x^4}{2}$ . Writing $\frac{x^2}{2}$ instead, as if the exponents did not compose, leaves the outer integrand as $\frac{x^4}{2}$ and the answer as $\frac{16}{5} = 3.2$ , roughly a third of the truth.

A different misreading treats the region as lying above the parabola. With no ceiling stated, that region is unbounded. The phrase under the curve, with no lower boundary given, always means down to the $x$ -axis.

Substitute the boundary curve into the antiderivative, not into the original integrand.
Check degrees: $x^2$ times a $y$ of size $x^2$ integrates to something of size $x^2 \cdot x^4 = x^6$ .
Keep powers of $2$ symbolic. Here $2^7 = 128$ , so the answer is $\frac{128}{14}$ before reducing.

The same region as type 2

Turn the slices horizontal. The region reaches up to $y = 4$ , and for a fixed $y$ the slice runs from the parabola $x = \sqrt{y}$ across to the vertical line $x = 2$ .

\int_0^4 \int_{\sqrt{y}}^{2} x^2 y\,dx\,dy = \int_0^4 y \cdot \frac{8 - y^{3/2}}{3}\,dy = \frac{1}{3}\left(64 - \frac{256}{7}\right) = \frac{64}{7}

Same answer, more work. The type 2 version costs a fractional power and a messier evaluation, which is the usual reason to prefer whichever order keeps the exponents whole.

Frequently asked questions

Why does a 7 appear in the denominator?

Because the inner integral raises the degree to $x^6$ , and integrating $x^6$ produces $\frac{x^7}{7}$ . The $7$ is a fingerprint of the parabolic top: over a rectangle the same integrand would leave only the powers the integrand itself supplies.

What is the area of this region, and the average value there?

The area is $\int_0^2 x^2\,dx = \frac{8}{3}$ . Dividing the integral by it gives the average of $x^2 y$ on the region, $\frac{64}{7} \cdot \frac{3}{8} = \frac{24}{7} \approx 3.43$ .