Chapter 4

$C\left(y\right)$, meaning that instead of being just a constant, like the number 5, it is a function of $y$, which acts like a constant when taking derivatives with respect to $x$.

curve to curve, then from point to point

[-2]

1. (a)

   $18x^2+42x-117$

2. (b)

   $-108$

[-2]

1. (a)

   $x^4/2-x^2+2x-3/2$

2. (b)

   $23/15$

[-2]

1. (a)

   ${\sin }^{2}y$

2. (b)

   $\pi /2$

${\displaystyle {\int }_1^4}{\displaystyle {\int }_{-2}^1}𝑑y𝑑x$ and ${\displaystyle {\int }_{-2}^1}{\displaystyle {\int }_1^4}𝑑x𝑑y$.

area of $R=9{\text{ units}}^2$

${\displaystyle {\int }_0^1}{\displaystyle {\int }_{x^4}^{\sqrt{x}}}𝑑y𝑑x$ and ${\displaystyle {\int }_0^1}{\displaystyle {\int }_{y^2}^{\sqrt[4]{y}}}𝑑x𝑑y$

area of $R=7/15{\text{ units}}^2$

[-2]

area of $R={\displaystyle {\int }_0^4}{\displaystyle {\int }_{-\sqrt{4-y}}^{\sqrt{4-y}}}𝑑x𝑑y$

[-2]

area of $R={\displaystyle {\int }_0^4}{\displaystyle {\int }_{-\sqrt{4-x^2/4}}^{\sqrt{4-x^2/4}}}𝑑y𝑑x$

[-2]

area of $R={\displaystyle {\int }_{-1}^2}{\displaystyle {\int }_{x^2}^{x+2}}𝑑y𝑑x$

volume

The double integral gives the signed volume under the surface. Since the surface is always positive, it is always above the $x$-$y$ plane and hence produces only “positive” volume.

6; ${\displaystyle {\int }_{-1}^1}{\displaystyle {\int }_1^2}\left({\displaystyle \frac{x}{y}}+3\right)𝑑y𝑑x$

112/3; ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^{4-2y}}\left(3x^2-y+2\right)𝑑x𝑑y$

16/5; ${\displaystyle {\int }_{-1}^1}{\displaystyle {\int }_0^{1-x^2}}\left(x+y+2\right)𝑑y𝑑x$

[-2]

1. (a)
2. (b)

   ${\displaystyle {\int }_0^1}{\displaystyle {\int }_{x^2}^{\sqrt{x}}}{x}^{2}y𝑑y𝑑x={\displaystyle {\int }_0^1}{\displaystyle {\int }_{y^2}^{\sqrt{y}}}{x}^{2}y𝑑x𝑑y$.

3. (c)

   $\frac{3}{56}$

[-2]

1. (a)
2. (b)

   ${\displaystyle {\int }_{-1}^1}{\displaystyle {\int }_{-1}^1}{x}^2-y^{2}dydx={\displaystyle {\int }_{-1}^1}{\displaystyle {\int }_{-1}^1}{x}^2-y^{2}dxdy$.

3. (c)

   0

[-2]

1. (a)
2. (b)
3. (c)

   ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^{3-3/2x}}\left(6-3x-2y\right)𝑑y𝑑x={\displaystyle {\int }_0^3}{\displaystyle {\int }_0^{2-2/3y}}\left(6-3x-2y\right)𝑑x𝑑y$.

4. (d)

   6

[-2]

1. (a)
2. (b)

   ${\displaystyle {\int }_{-3}^3}{\displaystyle {\int }_0^{\sqrt{9-x^2}}}\left(x^{3}y-x\right)𝑑y𝑑x={\displaystyle {\int }_0^3}{\displaystyle {\int }_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}}\left(x^{3}y-x\right)𝑑x𝑑y$.

3. (c)

   0

Integrating $e^{x^2}$ with respect to $x$ is not possible in terms of elementary functions. ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^{2x}}{e}^{x^2}𝑑y𝑑x=e^4-1$.

Integrating ${\displaystyle {\int }_y^1}{\displaystyle \frac{2y}{x^2+y^2}}𝑑x$ gives ${\tan }^{-1}\left(1/y\right)-\pi /4$; integrating ${\tan }^{-1}\left(1/y\right)$ is hard.

${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^x}{\displaystyle \frac{2y}{x^2+y^2}}𝑑y𝑑x=\ln 2$.

average value of $f=6/2=3$

average value of $f=\frac{112/3}{4}=28/3$

$f(r\cos \theta ,r\sin \theta )$, $rdrd\theta$

${\displaystyle {\int }_0^{2\pi }}{\displaystyle {\int }_0^1}\left(3r\cos \theta -r\sin \theta +4\right)r𝑑r𝑑\theta =4\pi$

${\displaystyle {\int }_0^{\pi }}{\displaystyle {\int }_{\cos \theta }^{3\cos \theta }}\left(8-r\sin \theta \right)r𝑑r𝑑\theta =16\pi$

${\displaystyle {\int }_0^{2\pi }}{\displaystyle {\int }_1^2}\left(\ln \left(r^2\right)\right)r𝑑r𝑑\theta =2\pi \left(\ln 16-3/2\right)$

${\displaystyle {\int }_{-\pi /2}^{\pi /2}}{\displaystyle {\int }_0^6}\left(r^2{\cos }^2\theta -r^2{\sin }^2\theta \right)r𝑑r𝑑\theta ={\displaystyle {\int }_{-\pi /2}^{\pi /2}}{\displaystyle {\int }_0^6}\left(r^2\cos \left(2\theta \right)\right)r𝑑r𝑑\theta =0$

${\displaystyle {\int }_{-\pi /2}^{\pi /2}}{\displaystyle {\int }_0^5}\left(r^2\right)𝑑r𝑑\theta =125\pi /3$

${\displaystyle {\int }_0^{\pi /4}}{\displaystyle {\int }_0^{\sqrt{8}}}\left(r\cos \theta +r\sin \theta \right)r𝑑r𝑑\theta =16\sqrt{2}/3$

[-2]

1. (a)

   This is impossible to integrate with rectangular coordinates as $e^{-\left(x^2+y^2\right)}$ does not have an antiderivative in terms of elementary functions.

2. (b)

   ${\displaystyle {\int }_0^{2\pi }}{\displaystyle {\int }_0^a}r{e}^{r^2}𝑑r𝑑\theta =\pi \left(1-e^{-a^2}\right)$.

3. (c)

   $\underset{a\to \mathrm{\infty }}{lim}\pi \left(1-e^{-a^2}\right)=\pi$. This implies that there is a finite volume under the surface $e^{-\left(x^2+y^2\right)}$ over the entire $x$-$y$ plane.

4. (d)

   If $R={ℝ}^2$, we can write the original integral as ${\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-t^2}𝑑t\right)}^2=\pi$.

$3\pi /4â\mathrm{ˆ}\mathrm{’}9\sqrt{3}/16$

$2$

$2\left(1â\mathrm{ˆ}\mathrm{’}{a}^3\right)/3\left(1â\mathrm{ˆ}\mathrm{’}{a}^2\right)$; $2/3$; $1$

Because they are scalar multiples of each other.

“little masses”

$M_x$ measures the moment about the $x$-axis, meaning we need to measure distance from the $x$-axis. Such measurements are measures in the $y$-direction.

$\overline{x}=5.25$

$(\overline{x},\overline{y})=(0,3)$

$M=150$g;

$M=2$lb

$M=16\pi \approx 50.27$kg

$M=54\pi \approx 169.65$lb

$M=150$g; $M_y=600$; $M_x=-75$; $(\overline{x},\overline{y})=(4,-0.5)$

$M=2$lb; $M_y=0$; $M_x=2/3$; $(\overline{x},\overline{y})=(0,1/3)$

$M=16\pi \approx 50.27$kg; $M_y=4\pi$; $M_x=4\pi$; $(\overline{x},\overline{y})=(1/4,1/4)$

$M=54\pi \approx 169.65$lb; $M_y=0$; $M_x=504$; $(\overline{x},\overline{y})=(0,2.97)$

$I_x=64/3$; $I_y=64/3$; $I_O=128/3$

$I_x=16/3$; $I_y=64/3$; $I_O=80/3$

arc length

surface areas

Intuitively, adding $h$ to $f$ only shifts $f$ up (i.e., parallel to the $z$-axis) and does not change its shape. Therefore it will not change the surface area over $R$.

Analytically, $f_x=g_x$ and $f_y=g_y$; therefore, the surface area of each is computed with identical double integrals.

$S={\displaystyle {\int }_0^{2\pi }}{\displaystyle {\int }_0^{2\pi }}\sqrt{1+{\cos }^{2}x{\cos }^{2}y+{\sin }^{2}x{\sin }^{2}y}𝑑x𝑑y$

$S={\displaystyle {\int }_{-1}^1}{\displaystyle {\int }_{-1}^1}\sqrt{1+4x^2+4y^2}𝑑x𝑑y$

$S={\displaystyle {\int }_0^3}{\displaystyle {\int }_{-1}^1}\sqrt{1+9+49}𝑑x𝑑y=6\sqrt{59}\approx 46.09$

[-2]

surface to surface, curve to curve and point to point

Answers can vary. From this section we used triple integration to find the volume of a solid region, the mass of a solid, and the center of mass of a solid.

$V={\int }_{-1}^1{\int }_{-1}^1\left(8-x^2-y^2-\left(2x+y\right)\right)𝑑x𝑑y=88/3$

$V={\int }_0^{\pi }{\int }_0^x\left(\cos x\sin y+2-\sin x\cos y\right)𝑑y𝑑x={\pi }^2-\pi \approx 6.728$

$dzdydx$: ${\displaystyle {\int }_0^3}{\displaystyle {\int }_0^{1-x/3}}{\displaystyle {\int }_0^{2-2x/3-2y}}𝑑z𝑑y𝑑x$

$dzdxdy$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{3-3y}}{\displaystyle {\int }_0^{2-2x/3-2y}}𝑑z𝑑x𝑑y$

$dydzdx$: ${\displaystyle {\int }_0^3}{\displaystyle {\int }_0^{2-2x/3}}{\displaystyle {\int }_0^{1-x/3-z/2}}𝑑y𝑑z𝑑x$

$dydxdz$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^{3-3z/2}}{\displaystyle {\int }_0^{1-x/3-z/2}}𝑑y𝑑x𝑑z$

$dxdzdy$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{2-2y}}{\displaystyle {\int }_0^{3-3y-3z/2}}𝑑x𝑑z𝑑y$

$dxdydz$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^{1-z/2}}{\displaystyle {\int }_0^{3-3y-3z/2}}𝑑x𝑑y𝑑z$

$V={\displaystyle {\int }_0^3}{\displaystyle {\int }_0^{1-x/3}}{\displaystyle {\int }_0^{2-2x/3-2y}}𝑑z𝑑y𝑑x=1.$

$dzdydx$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_{-2}^0}{\displaystyle {\int }_{y^2/2}^{-y}}𝑑z𝑑y𝑑x$

$dzdxdy$: ${\displaystyle {\int }_{-2}^0}{\displaystyle {\int }_0^2}{\displaystyle {\int }_{y^2/2}^{-y}}𝑑z𝑑x𝑑y$

$dydzdx$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^2}{\displaystyle {\int }_{-\sqrt{2z}}^{-z}}𝑑y𝑑z𝑑x$

$dydxdz$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^2}{\displaystyle {\int }_{-\sqrt{2z}}^{-z}}𝑑y𝑑x𝑑z$

$dxdzdy$: ${\displaystyle {\int }_{-2}^0}{\displaystyle {\int }_{y^2/2}^{-y}}{\displaystyle {\int }_0^2}𝑑x𝑑z𝑑y$

$dxdydz$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_{-\sqrt{2z}}^{-z}}{\displaystyle {\int }_0^2}𝑑x𝑑y𝑑z$

$V={\displaystyle {\int }_0^2}{\displaystyle {\int }_0^2}{\displaystyle {\int }_{-\sqrt{2z}}^{-z}}𝑑y𝑑z𝑑x=4/3.$

$dzdydx$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_{1-x/2}^1}{\displaystyle {\int }_0^{2x+4y-4}}𝑑z𝑑y𝑑x$

$dzdxdy$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_{2-2y}^2}{\displaystyle {\int }_0^{2x+4y-4}}𝑑z𝑑x𝑑y$

$dydzdx$: ${\displaystyle {\int }_0^2}{\displaystyle {\int }_0^{2x}}{\displaystyle {\int }_{z/4-x/2+1}^1}𝑑y𝑑z𝑑x$

$dydxdz$: ${\displaystyle {\int }_0^4}{\displaystyle {\int }_{z/2}^2}{\displaystyle {\int }_{z/4-x/2+1}^1}𝑑y𝑑x𝑑z$

$dxdzdy$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{4y}}{\displaystyle {\int }_{z/2-2y+2}^2}𝑑x𝑑z𝑑y$

$dxdydz$: ${\displaystyle {\int }_0^4}{\displaystyle {\int }_{z/4}^1}{\displaystyle {\int }_{z/2-2y+2}^2}𝑑x𝑑y𝑑z$

$V={\displaystyle {\int }_0^4}{\displaystyle {\int }_{z/4}^1}{\displaystyle {\int }_{z/2-2y+2}^2}𝑑x𝑑y𝑑z=4/3.$

$dzdydx$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{1-x^2}}{\displaystyle {\int }_0^{\sqrt{1-y}}}𝑑z𝑑y𝑑x$

$dzdxdy$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{\sqrt{1-y}}}{\displaystyle {\int }_0^{\sqrt{1-y}}}𝑑z𝑑x𝑑y$

$dydzdx$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^x}{\displaystyle {\int }_0^{1-x^2}}𝑑y𝑑z𝑑x+{\displaystyle {\int }_0^1}{\displaystyle {\int }_x^1}{\displaystyle {\int }_0^{1-z^2}}𝑑y𝑑z𝑑x$

$dydxdz$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^z}{\displaystyle {\int }_0^{1-z^2}}𝑑y𝑑x𝑑z+{\displaystyle {\int }_0^1}{\displaystyle {\int }_z^1}{\displaystyle {\int }_0^{1-x^2}}𝑑y𝑑x𝑑z$

$dxdzdy$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{\sqrt{1-y}}}{\displaystyle {\int }_0^{\sqrt{1-y}}}𝑑x𝑑z𝑑y$

$dxdydz$: ${\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{1-z^2}}{\displaystyle {\int }_0^{\sqrt{1-y}}}𝑑x𝑑y𝑑z$

Answers will vary. Neither order is particularly “hard.” The order $dzdydx$ requires integrating a square root, so powers can be messy; the order $dydzdx$ requires two triple integrals, but each uses only polynomials.

8

$\pi$

$M=10$, $M_{yz}=15/2$, $M_{xz}=5/2$, $M_{xy}=5$;

$(\overline{x},\overline{y},\overline{z})=(3/4,1/4,1/2)$

$M=16/5$, $M_{yz}=16/3$, $M_{xz}=104/45$, $M_{xy}=32/9$;

$(\overline{x},\overline{y},\overline{z})=(5/3,13/18,10/9)\approx (1.67,0.72,1.11)$

In cylindrical, $r$ determines how far from the origin one goes in the $x$-$y$ plane before considering the $z$-component. Equivalently, if on projects a point in cylindrical coordinates onto the $x$-$y$ plane, $r$ will be the distance of this projection from the origin.

In spherical, $\rho$ is the distance from the origin to the point.

Cylinders (tubes) centered at the origin, parallel to the $z$-axis; planes parallel to the $z$-axis that intersect the $z$-axis; planes parallel to the $x$-$y$ plane.

[-2]

1. (a)

   Cylindrical: $(2\sqrt{2},\pi /4,1)$ and $(2,5\pi /6,0)$

   Spherical: $(3,\pi /4,{\cos }^{-1}\left(1/3\right))$ and $(2,5\pi /6,\pi /2)$

2. (b)

   Rectangular: $(\sqrt{2},\sqrt{2},2)$ and $(0,-3,-4)$

   Spherical: $(2\sqrt{2},\pi /4,\pi /4)$ and $(5,3\pi /2,\pi -{\tan }^{-1}\left(3/4\right))$

3. (c)

   Rectangular: $(1,1,\sqrt{2})$ and $(0,0,1)$

   Cylindrical: $(\sqrt{2},\pi /4,\sqrt{2})$ and $(0,0,1)$

[-2]

1. (a)

   A cylindrical surface or tube, centered along the $z$-axis of radius 1, extending from the $x$-$y$ plane up to the plane $z=1$ (i.e., the tube has a length of 1).

2. (b)

   This is a region of space, being half of a tube with “thick” walls of inner radius 1 and outer radius 2, centered along the $z$-axis with a length of 1, where the half “below” the $x$-$z$ plane is removed.

3. (c)

   This is upper half of the sphere of radius 3 centered at the origin (i.e., the upper hemisphere).

4. (d)

   This is a region of space, where the ball of radius 2, centered at the origin, is removed from the ball of radius 3, centered at the origin.

${\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}}{\displaystyle {\int }_{r_1}^{r_2}}{\displaystyle {\int }_{z_1}^{z_2}}h(r,\theta ,z)r𝑑z𝑑r𝑑\theta$

The region in space is bounded between the planes $z=0$ and $z=2$, inside of the cylinder $x^2+y^2=4$, and the planes $\theta =0$ and $\theta =\pi /2$: describes a “wedge” of a cylinder of height 2 and radius 2; the angle of the wedge is $\pi /2$, or ${90}^{\circ }$.

Bounded between the plane $z=1$ and the cone $z=1-\sqrt{x^2+y^2}$: describes an inverted cone, with height of 1, point at $(0,0,1)$ and base radius of 1.

Describes a quarter of a ball of radius 3, centered at the origin; the quarter resides above the $x$-$y$ plane and above the $x$-$z$ plane.

Describes the portion of the unit ball that resides in the first octant.

Bounded above the cone $z=\sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=4$: describes a shape that is somewhat “diamond”-like; some think of it as looking like an ice cream cone (see Figure 4.56). It describes a cone, where the side makes an angle of $\pi /4$ with the positive $z$-axis, topped by the portion of the ball of radius 2, centered at the origin.

The region in space is bounded below by the cone $z=\sqrt{3}\sqrt{x^2+y^2}$ and above by the plane $z=1$: it describes a cone, with point at the origin, centered along the positive $z$-axis, with height of 1 and base radius of $\tan \left(\pi /6\right)=1/\sqrt{3}$.

We find $M_{yz}=0$, $M_{xz}=0$, and $M_{xy}=224\pi /3$, placing the center of mass at $(0,0,2)$.

We find $M_{yz}=0$, $M_{xz}=8/3-\pi /8$, and $M_{xy}=65\pi /16-8/3$, placing the center of mass at $\approx (0,0.405,1.80)$.

We find $M_{yz}=0$, $M_{xz}=0$, and $M_{xy}=\pi /4$, placing the center of mass at $(0,0,3/8)$.

We find $M_{yz}=0$, $M_{xz}=0$, and $M_{xy}=\left(4-\sqrt{2}\right)\pi /30$, placing the center of mass at $(0,0,4\left(4-\sqrt{2}\right)/15)$.