Multivariable calculus

Double Integral of x^3 y^2 Over [-1,2] by [0,2]

The double integral of x cubed times y squared over the rectangle x from -1 to 2 and y from 0 to 2 equals 10. The integrand factors: the integral of x cubed from -1 to 2 is 15/4, the integral of y squared from 0 to 2 is 8/3, and 15/4 times 8/3 is 10.

f(x,y)=x3y2f(x,y) = x^3 y^2
∫−12∫02x3y2 dy dx=10\int_{-1}^{2}\int_{0}^{2} x^3 y^2\,dy\,dx = 10

Numerically 10.000000, confirmed by quadrature on every build.

Factor, and watch the negative limit

The integrand is a product with one variable per factor and the limits are constants, so the double integral is a product of two single integrals. The only delicate step is the lower limit x=−1x = -1.

∫−12x3 dx=[x44]−12=164−14=154\int_{-1}^{2} x^3\,dx = \left[\frac{x^4}{4}\right]_{-1}^{2} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}

Note (−1)4=1(-1)^4 = 1, not −1-1. The antiderivative carries the fourth power even though the integrand carries the third, and that change of parity is where sign errors get in.

∫02y2 dy=83,154⋅83=10\int_{0}^{2} y^2\,dy = \frac{8}{3}, \qquad \frac{15}{4}\cdot\frac{8}{3} = 10

The answer is a signed volume

On the strip where −1≤x<0-1 \le x < 0 the integrand is negative, since x3<0x^3 < 0 while y2≥0y^2 \ge 0. That part of the rectangle subtracts from the total instead of adding to it.

∫−10∫02x3y2 dy dx=−14⋅83=−23,∫02∫02x3y2 dy dx=4⋅83=323\int_{-1}^{0}\int_{0}^{2} x^3 y^2\,dy\,dx = -\frac{1}{4}\cdot\frac{8}{3} = -\frac{2}{3}, \qquad \int_{0}^{2}\int_{0}^{2} x^3 y^2\,dy\,dx = 4\cdot\frac{8}{3} = \frac{32}{3}

The two pieces add to 323−23=10\frac{32}{3} - \frac{2}{3} = 10. If you wanted the volume between the surface and the plane with no cancellation, you would integrate ∣x3y2∣\left|x^3 y^2\right| instead and get 323+23=343≈11.33\frac{32}{3} + \frac{2}{3} = \frac{34}{3} \approx 11.33.

The mistake: sign slips at the lower limit

[x44]−12\left[\frac{x^4}{4}\right]_{-1}^{2} means 244−(−1)44\frac{2^4}{4} - \frac{(-1)^4}{4}. Adding instead of subtracting gives 174\frac{17}{4} and turns the answer into 343≈11.33\frac{34}{3} \approx 11.33.

That is not a coincidence. Flipping the sign of the lower limit's contribution counts the negative strip as positive, which is exactly the unsigned volume from the previous section, so the wrong answer looks respectable.

  • Evaluate the antiderivative at the top limit minus at the bottom limit, in that order, every time.
  • An even power of a negative number is positive: (−1)4=1(-1)^4 = 1.
  • Since the integrand factors, confirm the two single integrals separately: 154=3.75\frac{15}{4} = 3.75 and 83≈2.667\frac{8}{3} \approx 2.667 multiply to 10.

Frequently asked questions

Why is the answer positive when the integrand is negative on part of the rectangle?

Because the positive part dominates. The strip with x>0x > 0 contributes 323\frac{32}{3} and the strip with x<0x < 0 contributes −23-\frac{2}{3}, so the signed total is 10.

What would the integral be over the rectangle from -2 to 2 in x?

Exactly 0. On a symmetric interval ∫−22x3 dx=0\int_{-2}^{2} x^3\,dx = 0 because x3x^3 is odd, and a factor of 0 wipes out the product no matter what the yy integral is.