Multivariable calculus

Double Integral of y sin x Under a Cosine Arch

Q: Could this be set up as a type 2 region instead?

Yes, with y from 0 to 1 and x from 0 to \arccos y, since the curve inverts cleanly on this interval. The \arccos then disappears: the inner integral is y\Big[-\cos x\Big]_{0}^{\arccos y} = y(1-y), and \int_{0}^{1}\left(y - y^{2}\right)\,dy = \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}. It is a clean independent check on the answer.

Q: Why does the answer come out so tidy?

Because the outer integral reduces to \int_{0}^{1} u^2\,du after substitution. The trigonometric functions cancel out of the final expression entirely, leaving a rational number.

The double integral of y times sin(x) over the region where x runs from 0 to pi/2 and y runs from 0 up to cos(x) equals 1/6, about 0.1667. The inner integral in y gives one half of sin(x) times cos(x) squared, and the substitution u = cos(x) finishes it.

f(x,y) = y\sin x

\dfrac{1}{6}

Numerically 0.166667, confirmed by quadrature on every build.

A type 1 region: y between two functions of x

The region sits between the $x$ axis below and the curve $y = \cos x$ above, for $x$ from $0$ to $\pi/2$ . On a type 1 region the inner variable is $y$ and its bounds are functions of $x$ .

\iint_{D} y \sin x\,dA = \int_{0}^{\pi/2}\!\!\int_{0}^{\cos x} y \sin x\,dy\,dx

Integrate in $y$ with $x$ held fixed. Only the $y$ matters inside; $\sin x$ rides along as a constant.

\int_{0}^{\cos x} y \sin x\,dy = \sin x\left[\frac{y^{2}}{2}\right]_{0}^{\cos x} = \frac{\sin x\cos^{2}x}{2}

Finish with the substitution u = cos x

The outer integral now has $\sin x$ sitting next to a power of $\cos x$ , which is the signature of the substitution $u = \cos x$ , $du = -\sin x\,dx$ .

\frac{1}{2}\int_{0}^{\pi/2} \sin x\cos^{2}x\,dx = \frac{1}{2}\int_{1}^{0} u^{2}\,(-du) = \frac{1}{2}\int_{0}^{1} u^{2}\,du = \frac{1}{6}

Notice the limits changed with the substitution: $x = 0$ gives $u = 1$ , and $x = \pi/2$ gives $u = 0$ . Flipping them absorbs the minus sign from $du$ .

\iint_{D} y \sin x\,dA = \frac{1}{6} \approx 0.1667

The mistake: using a constant ceiling instead of the curve

Since $\cos x$ never exceeds $1$ on this interval, students sometimes replace the upper bound by the constant $1$ and integrate over the rectangle $[0,\pi/2]\times[0,1]$ instead.

\int_{0}^{\pi/2}\!\!\int_{0}^{1} y \sin x\,dy\,dx = \frac{1}{2}\int_{0}^{\pi/2}\sin x\,dx = \frac{1}{2}

That is $0.5$ against the true $0.1667$ , three times too large, because the rectangle includes everything above the cosine curve. The bound $\cos x$ is not an estimate of the region, it is the region's boundary.

The tell that a bound must stay variable: the answer to the inner integral should still contain $x$ . If your inner integral comes out as a pure number on a type 1 region, you have flattened the boundary.

Frequently asked questions

Could this be set up as a type 2 region instead?

Yes, with $y$ from $0$ to $1$ and $x$ from $0$ to $\arccos y$ , since the curve inverts cleanly on this interval. The $\arccos$ then disappears: the inner integral is $y\Big[-\cos x\Big]_{0}^{\arccos y} = y(1-y)$ , and $\int_{0}^{1}\left(y - y^{2}\right)\,dy = \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}$ . It is a clean independent check on the answer.

Why does the answer come out so tidy?

Because the outer integral reduces to $\int_{0}^{1} u^2\,du$ after substitution. The trigonometric functions cancel out of the final expression entirely, leaving a rational number.