Multivariable calculus

Double Integral of y sin(xy) Over a Rectangle

Q: Is 1 - 2/pi a plausible size for this volume?

Yes. The integrand y\sin(xy) never exceeds 1 on this region and is 0 along y = 0, so the volume must be well under the base area \pi/2 \approx 1.571. The value 0.3634 fits.

The double integral of y times sin(xy), with x from 0 to pi/2 and y from 0 to 1, equals 1 - 2/pi, about 0.3634. Integrating in x first is the easy order: the inner integral is a plain substitution and gives 1 - cos(pi y / 2).

f(x,y) = y\sin(xy)

1 - \dfrac{2}{\pi}

Numerically 0.363380, confirmed by quadrature on every build.

Integrate in x first, where the y out front is the chain-rule factor

Hold $y$ fixed. Then $y\sin(xy)$ is exactly the $x$ derivative of $-\cos(xy)$ , so the inner integral needs no technique beyond recognising it.

\int_{0}^{\pi/2} y\sin(xy)\,dx = \Big[-\cos(xy)\Big]_{x=0}^{x=\pi/2} = 1 - \cos\!\left(\frac{\pi y}{2}\right)

Now integrate that in $y$ over $[0,1]$ . The cosine picks up a factor of $2/\pi$ when you undo its inner derivative.

\int_{0}^{1}\left(1 - \cos\!\left(\frac{\pi y}{2}\right)\right)\,dy = \left[y - \frac{2}{\pi}\sin\!\left(\frac{\pi y}{2}\right)\right]_{0}^{1} = 1 - \frac{2}{\pi}

Numerically $1 - 2/\pi \approx 0.3634$ .

Why the other order needs integration by parts

Fix $x$ instead and integrate in $y$ . Now the factor $y$ is a genuine second function, not a chain-rule leftover, so parts is unavoidable.

\int_{0}^{1} y\sin(xy)\,dy = \frac{\sin x}{x^{2}} - \frac{\cos x}{x}

That is a detour, not a dead end. The expression is the derivative of $-\frac{\sin x}{x}$ , so the outer integral does close in elementary terms once you spot that.

\int_{0}^{\pi/2}\left(\frac{\sin x}{x^{2}} - \frac{\cos x}{x}\right)\,dx = \left[-\frac{\sin x}{x}\right]_{0}^{\pi/2} = -\frac{2}{\pi} + 1

Same number, three extra moves: parts, a removable singularity at $x = 0$ where $\frac{\sin x}{x} \to 1$ , and a quotient-rule derivative read backwards. The skill being tested is choosing the order before you start, not surviving the wrong one.

Look at which variable appears as a bare multiplying factor. Here it is $y$ .
That factor is usually the chain-rule derivative of the inner function, so integrate with respect to the other variable, $x$ .
The rule of thumb: integrate first in the variable that is not the loose factor out front.

The mistake: pulling y out and forgetting it is inside the sine

Because $y$ is constant during the $x$ integration, it is tempting to write $y\int_{0}^{\pi/2}\sin(xy)\,dx$ and then integrate $\sin(xy)$ as though it were $\sin x$ .

\int \sin(xy)\,dx = -\frac{\cos(xy)}{y}, \qquad \text{not} \quad -\cos(xy)

The missing $1/y$ is exactly what cancels the $y$ out front. Dropping it leaves an answer that still has a stray $y$ in it, and the outer integral then produces a different number.

Check any antiderivative by differentiating it back. If $\frac{\partial}{\partial x}$ of your inner answer does not reproduce $y\sin(xy)$ exactly, the chain-rule factor is wrong.

Frequently asked questions

Why does the inner variable here have the constant bounds?

The region is a rectangle, so both variables have constant bounds and either can be inner. The order was chosen for convenience, not necessity. With $x$ inner from $0$ to $\pi/2$ , the inner integral is a one-line substitution.

Is 1 - 2/pi a plausible size for this volume?

Yes. The integrand $y\sin(xy)$ never exceeds $1$ on this region and is $0$ along $y = 0$ , so the volume must be well under the base area $\pi/2 \approx 1.571$ . The value $0.3634$ fits.