Triple integrals: iterated limits, cylindrical and spherical coordinates

Worked solutions

Problem 1

1. $\displaystyle\iiint_B xyz\,dV = \int_0^1 x\,dx\int_0^2 y\,dy\int_0^3 z\,dz$The integrand is a product of one function of each variable and the limits are all numbers, so the integral splits.
2. $\int_0^1 x\,dx = \tfrac12$, $\int_0^2 y\,dy = 2$, $\int_0^3 z\,dz = \tfrac92$Three single integrals.
3. $\tfrac12\cdot2\cdot\tfrac92 = \tfrac92$Multiply.
4. Iterated directly: $\int_0^3 z\,dz = \tfrac92$, then $\int_0^2 \tfrac92 y\,dy = 9$, then $\int_0^1 9x\,dx = \tfrac92$The same number, innermost first, as a check.
5. $\tfrac92 = 4.5$Splitting works only when both the integrand and the region are products; over a tetrahedron the limits tie the variables together.

Problem 2

1. $z$ runs from $0$ up to $1 - x - y$; the shadow on the $xy$-plane is the triangle $x, y \ge 0$, $x + y \le 1$A vertical line through $(x, y)$ enters $T$ at the floor and leaves through the slanted face.
2. $V = \int_0^1\int_0^{1-x}(1 - x - y)\,dy\,dx = \int_0^1 \dfrac{(1 - x)^2}{2}\,dx = \dfrac16$The inner $y$-integral of $c - y$ from $0$ to $c$, with $c = 1 - x$, is $\tfrac{c^2}{2}$.
3. $\iiint_T x\,dV = \int_0^1 x\,\dfrac{(1 - x)^2}{2}\,dx = \dfrac12\Big(\dfrac12 - \dfrac23 + \dfrac14\Big) = \dfrac1{24}$The same inner integrals, with the extra factor $x$ carried along.
4. $\bar x = \dfrac{1/24}{1/6} = \dfrac14$Divide by the volume.
5. $V = \tfrac16$ and $\bar x = \tfrac14$By symmetry the centroid is $\big(\tfrac14, \tfrac14, \tfrac14\big)$: a quarter of the way from the base to the opposite vertex, as for any tetrahedron.

Problem 3

1. $0 \le z \le 6 - 2x - 3y$Floor $z = 0$, roof the plane solved for $z$.
2. $0 \le y \le \dfrac{6 - 2x}{3}$The shadow is where the roof is above the floor: $2x + 3y \le 6$ with $x, y \ge 0$. A line in the $y$-direction leaves it at $3y = 6 - 2x$.
3. $0 \le x \le 3$The shadow's extent in $x$, where $2x = 6$ on the $x$-axis. Only this outer range is a pair of numbers.
4. $\displaystyle\int_0^3\int_0^{(6-2x)/3}(6 - 2x - 3y)\,dy\,dx = \int_0^3\frac{(6 - 2x)^2}{6}\,dx = 6$With $c = 6 - 2x$, the inner integral of $c - 3y$ from $0$ to $\tfrac c3$ is $\tfrac{c^2}{3} - \tfrac{c^2}{6} = \tfrac{c^2}{6}$; then $\int_0^3 \tfrac{(6 - 2x)^2}{6}\,dx = \tfrac{1}{6}\cdot\tfrac{216}{6} = 6$.
5. $V = 6$Check with the formula for a tetrahedron, $\tfrac16$ of the product of the intercepts: $\tfrac16\cdot3\cdot2\cdot6 = 6$. Taking the limits as $0$ to $6$, $0$ to $2$ and $0$ to $3$ gives the box around it, $36$ (Mistake 4).

Problem 4

1. $r^2 \le z \le 4$, $0 \le r \le 2$, $0 \le \theta \le 2\pi$The paraboloid is $z = r^2$; it meets $z = 4$ at $r = 2$.
2. $V = \int_0^{2\pi}\int_0^2\int_{r^2}^4 r\,dz\,dr\,d\theta = 2\pi\int_0^2(4 - r^2)\,r\,dr$The factor $r$ is part of $dV$; nothing in the integrand depends on $\theta$.
3. $2\pi\Big[2r^2 - \dfrac{r^4}{4}\Big]_0^2 = 2\pi(8 - 4) = 8\pi$Leaving out the $r$ gives $\tfrac{32\pi}{3}$ (Mistake 1).
4. $\iiint_E z\,dV = 2\pi\int_0^2 \dfrac{16 - r^4}{2}\,r\,dr = \pi\Big(32 - \dfrac{32}{3}\Big) = \dfrac{64\pi}{3}$The inner integral is $\int_{r^2}^4 z\,dz = \tfrac{16 - r^4}{2}$.
5. $V = 8\pi \approx 25.13$ and $\bar z = \dfrac{64\pi/3}{8\pi} = \dfrac83$The centroid is above the middle height $2$ because the solid is wide at the top and narrows to a point at the bottom.

Problem 5

1. $x^2 + y^2 = r^2$ and $dV = r\,dz\,dr\,d\theta$, with $0 \le z \le 3$, $0 \le r \le 2$In cylindrical coordinates the cylinder is a box.
2. $I_z = \int_0^{2\pi}\int_0^2\int_0^3 r^2\cdot r\,dz\,dr\,d\theta$The integrand $r^2$ times the $r$ from $dV$ gives $r^3$.
3. $= 2\pi\cdot\dfrac{2^4}{4}\cdot3 = 24\pi$The integral splits: $\int_0^{2\pi}d\theta = 2\pi$, $\int_0^2 r^3\,dr = 4$, $\int_0^3 dz = 3$.
4. The mass is $M = \pi\cdot2^2\cdot3 = 12\pi$, so $I_z = \tfrac12 M R^2$ with $R = 2$The textbook formula for a solid cylinder, as a check.
5. $I_z = 24\pi \approx 75.40$Mass far from the axis counts with the square of its distance, which is why a hollow cylinder of the same mass has twice this moment.

Problem 6

1. $0 \le \rho \le a$, $0 \le \varphi \le \pi$, $0 \le \theta \le 2\pi$$\varphi$ only needs to reach $\pi$, the negative $z$-axis; $\theta$ supplies the full turn.
2. $V = \int_0^{2\pi}\int_0^\pi\int_0^a \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$$dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.
3. $\int_0^a\rho^2\,d\rho = \dfrac{a^3}{3}$, $\int_0^\pi\sin\varphi\,d\varphi = 2$, $\int_0^{2\pi}d\theta = 2\pi$The integrand is a product and the limits are numbers, so it splits.
4. $V = 2\pi\cdot2\cdot\dfrac{a^3}{3}$Multiply.
5. $V = \tfrac43\pi a^3$Without the $\sin\varphi$ the result is $\tfrac23\pi^2 a^3 \approx 6.58a^3$ (Mistake 2), and running $\varphi$ to $2\pi$ gives $0$ (Mistake 3).

Problem 7

1. Spherical: $0 \le \rho \le 2$, $0 \le \varphi \le \tfrac\pi4$, $0 \le \theta \le 2\pi$The cone makes the angle $\tfrac\pi4$ with the $z$-axis, and the solid is the part of the ball within that angle.
2. $V = 2\pi\Big[-\cos\varphi\Big]_0^{\pi/4}\cdot\dfrac{2^3}{3} = 2\pi\Big(1 - \dfrac{\sqrt2}{2}\Big)\dfrac83$The integral splits as in Problem 6.
3. Cylindrical: $r \le z \le \sqrt{4 - r^2}$, and the cone meets the sphere where $r^2 + r^2 = 4$, at $r = \sqrt2$So $0 \le r \le \sqrt2$, not $2$.
4. $2\pi\int_0^{\sqrt2}\big(\sqrt{4 - r^2} - r\big)\,r\,dr = 2\pi\Big(\dfrac{8 - 2\sqrt2}{3} - \dfrac{2\sqrt2}{3}\Big)$$\int_0^{\sqrt2} r\sqrt{4 - r^2}\,dr = \Big[-\tfrac13(4 - r^2)^{3/2}\Big]_0^{\sqrt2} = \tfrac{8 - 2\sqrt2}{3}$ and $\int_0^{\sqrt2} r^2\,dr = \tfrac{2\sqrt2}{3}$.
5. $V = \dfrac{8\pi(2 - \sqrt2)}{3} \approx 4.907$Both routes agree. Spherical coordinates turn the solid into a box, which is why the first route is one line. Running $r$ to $2$ in the second route gives $0$ (Mistake 5).

Problem 8

1. $x^2 + y^2 + z^2 = \rho^2$The integrand depends only on the distance from the origin.
2. $\int_0^{2\pi}\int_0^\pi\int_0^1 \rho^2\cdot\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta = 2\pi\cdot2\cdot\dfrac15$$\int_0^1\rho^4\,d\rho = \tfrac15$.
3. $\iiint_B (x^2 + y^2 + z^2)\,dV = \dfrac{4\pi}{5}$Multiply.
4. $\iiint_B x^2\,dV = \iiint_B y^2\,dV = \iiint_B z^2\,dV$The ball looks the same after relabeling the axes, so the three integrals are equal and add to $\tfrac{4\pi}{5}$.
5. $\iiint_B \rho^2\,dV = \tfrac{4\pi}{5}$ and $\iiint_B z^2\,dV = \tfrac{4\pi}{15}$Directly: $\int_0^{2\pi}\int_0^\pi\int_0^1 \rho^4\cos^2\varphi\sin\varphi\,d\rho\,d\varphi\,d\theta = 2\pi\cdot\tfrac23\cdot\tfrac15 = \tfrac{4\pi}{15}$. Dividing by the volume $\tfrac{4\pi}{3}$, the average of $\rho^2$ over the ball is $\tfrac35$.

Problem 9

1. As written the inner integral $\int_y^1 e^{x^3}\,dx$ has no elementary antiderivativeSo change the order.
2. The region is $0 \le z \le y \le x \le 1$Read the inequalities off the limits: $z \le y$ from the middle, $y \le x$ from the inner, and all of them in $[0, 1]$.
3. In the order $dz\,dy\,dx$: $0 \le z \le y$, $0 \le y \le x$, $0 \le x \le 1$For fixed $x$ and $y$, $z$ runs up to $y$; for fixed $x$, $y$ runs up to $x$.
4. $\displaystyle\int_0^1\int_0^x\int_0^y e^{x^3}\,dz\,dy\,dx = \int_0^1 e^{x^3}\int_0^x y\,dy\,dx = \int_0^1 \frac{x^2}{2}e^{x^3}\,dx$$e^{x^3}$ is constant in the two inner integrals.
5. $\dfrac{e - 1}{6} \approx 0.2864$With $u = x^3$, $\int_0^1 \tfrac{x^2}{2}e^{x^3}\,dx = \tfrac16\int_0^1 e^u\,du$. The factor $\tfrac{x^2}{2}$ is the area of the triangle $0 \le z \le y \le x$ in the $yz$-plane, and it is exactly what the substitution needed.

Problem 10

1. $P(\|X\| \le 1) = (2\pi)^{-3/2}\displaystyle\int_0^{2\pi}\int_0^\pi\int_0^1 e^{-\rho^2/2}\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$The density depends only on $\rho$, so spherical coordinates turn the ball into a box.
2. $= (2\pi)^{-3/2}\cdot4\pi\displaystyle\int_0^1 \rho^2e^{-\rho^2/2}\,d\rho = \sqrt{\frac{2}{\pi}}\int_0^1 \rho^2e^{-\rho^2/2}\,d\rho$The angles give $4\pi$, the area of the unit sphere, and $4\pi(2\pi)^{-3/2} = \sqrt{2/\pi}$.
3. $\int_0^1 \rho^2e^{-\rho^2/2}\,d\rho = \Big[-\rho e^{-\rho^2/2}\Big]_0^1 + \int_0^1 e^{-\rho^2/2}\,d\rho = -e^{-1/2} + \sqrt{2\pi}\Big(\Phi(1) - \dfrac12\Big)$Parts with $u = \rho$ and $dv = \rho e^{-\rho^2/2}\,d\rho$; the last integral is a normal probability.
4. $P = 2\Phi(1) - 1 - \sqrt{\dfrac2\pi}\,e^{-1/2} \approx 0.6827 - 0.4839$Multiply through by $\sqrt{2/\pi}$.
5. $P(\|X\| \le 1) \approx 0.1987$Each coordinate is within $1$ of $0$ with probability $0.6827$, yet the vector is within distance $1$ less than $20\%$ of the time. In $d$ dimensions $\|X\|$ concentrates near $\sqrt d$, so a high-dimensional Gaussian has almost no mass near its mode.