Multivariable calculus

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

The double integral of y over the region bounded by the line x = y and the parabola x = 2 - y^2 equals -9/4, or -2.25. The curves meet at y = -2 and y = 1, and the parabola is the right-hand boundary between them. The value is negative because most of the region lies below the x-axis.

f(x,y)=yf(x,y) = y
∫−21∫y2−y2y dx dy=−94\int_{-2}^{1} \int_{y}^{2-y^2} y\,dx\,dy = -\frac{9}{4}

Numerically -2.250000, confirmed by quadrature on every build.

Type 2 keeps this to a single integral

Find the intersections in the outer variable: y=2−y2y = 2 - y^2 gives y2+y−2=(y+2)(y−1)=0y^2 + y - 2 = (y+2)(y-1) = 0, so y=−2y = -2 and y=1y = 1. Those are the outer limits. Between them test y=0y = 0: the line gives x=0x = 0 and the parabola gives x=2x = 2, so the parabola is the right-hand boundary.

∫y2−y2y dx=y(2−y2−y)=2y−y3−y2\int_{y}^{2-y^2} y\,dx = y\left(2 - y^2 - y\right) = 2y - y^3 - y^2
∫−21(2y−y3−y2) dy=[y2−y44−y33]−21=512−83=−94\int_{-2}^{1}\left(2y - y^3 - y^2\right)\,dy = \left[y^2 - \frac{y^4}{4} - \frac{y^3}{3}\right]_{-2}^{1} = \frac{5}{12} - \frac{8}{3} = -\frac{9}{4}

The lower endpoint is where sign errors breed. At y=−2y = -2 the even powers give 4−4=04 - 4 = 0, while the odd one flips: −y33-\frac{y^3}{3} becomes +83+\frac{8}{3}. So the bracket is 83\frac{8}{3}, and subtracting that positive number from the 512\frac{5}{12} at the top is exactly what drives the answer negative. Miss the flip and you get 512+83\frac{5}{12} + \frac{8}{3}, a positive 3712\frac{37}{12}.

The mistake: forcing it into type 1 and forgetting to split

Slice this region vertically and the description changes partway across. For xx between −2-2 and 11 the slice runs from the lower branch of the parabola up to the line. For xx between 11 and 22 it runs between the two branches of the parabola.

∫−21∫−2−xxy dy dx+∫12∫−2−x2−xy dy dx=−94\int_{-2}^{1}\int_{-\sqrt{2-x}}^{x} y\,dy\,dx + \int_{1}^{2}\int_{-\sqrt{2-x}}^{\sqrt{2-x}} y\,dy\,dx = -\frac{9}{4}

That is two integrals, a square root in the limits, and two more chances to lose a sign, against one clean polynomial in the type 2 order. The second piece is 00 by symmetry, but you only know that after setting it up.

  • Boundaries written as x=g(y)x = g(y) point to type 2.
  • A boundary whose formula changes partway means the other order needs a split.
  • Outer limits come from the intersections, solved in the outer variable.

Why the answer is negative

The integrand yy is negative on most of the region, which stretches from y=−2y = -2 up to only y=1y = 1. Nothing is wrong with a negative double integral: it is a signed sum, and only ∬1 dA\iint 1\,dA is forced to be positive.

area=∫−21(2−y2−y) dy=92,yˉ=−9/49/2=−12\text{area} = \int_{-2}^{1}\left(2 - y^2 - y\right)\,dy = \frac{9}{2}, \qquad \bar{y} = \frac{-9/4}{9/2} = -\frac{1}{2}

So the centroid sits at height −12-\frac12, below the xx-axis, which matches the picture: the region hangs a good deal further below the axis than above it.

Frequently asked questions

How do I get the outer limits -2 and 1?

Set the two boundary curves equal in the outer variable. Here y=2−y2y = 2 - y^2 becomes y2+y−2=(y+2)(y−1)=0y^2 + y - 2 = (y+2)(y-1) = 0. Solve rather than read them off a sketch; the algebra is exact and takes one line.

Would the answer be positive if I swapped the inner limits?

It would read +94+\frac94, and it would be wrong. The inner lower limit is the left boundary, here the line x=yx = y, and the upper limit is the right boundary, the parabola. Reversing them multiplies the integral by −1-1.