# Triple Integrals



My notes of the excellent lectures 25 and 26 by “Denis Auroux. 18.02 Multivariable Calculus. Fall 2007. Massachusetts Institute of Technology: MIT OpenCourseWare, https://ocw.mit.edu. License: Creative Commons BY-NC-SA.”

Recall: using a double integral of a you can find the volume between a certain region $$R$$ in the $$xy$$-plane, and a function $$f(x,y)$$.

$$\iint_R f(x,y)\,dA \nonumber$$

Using triple integrals we can find volume between two surfaces. We can even find the mass of a 3D object, when the volume of the region we’re interested in has a variable density.

## Definition

Triple integrals are just like double integrals, but in three dimensions. This extra dimension let’s us express the region $$R$$ in $$xyz$$-space.

$$\shaded{ \iiint_R f(x,y,z)\,dV } \nonumber$$

Here

• The shape of the region $$R$$ is described in the integrals’ bounds.
• When calculating mass, a non-uniform density will be described by function $$f$$ as $$\delta(x,y,z)$$.

Triple integrals are evaluated as three embedded integrals. The bounds of the inner integrals might be functions of the outer variables. $$\int_{z_{min}}^{z_{max}} \underbrace{ \int_{y_{min}(z)}^{y_{max}(z)} \overbrace{ \int_{x_{min}(y,z)}^{x_{max}(y,z)} f(x,y,z)\,dx }^{\text{function of only }y,z} \,dy }_{\text{function of only }z} \,dz \nonumber$$

## In Cartesian coordinates

### Volume element

The volume element for Cartesian coordinates is $$\shaded{ dV=dx\,dy\,dz } \nonumber$$

### Examples

#### One

Find the region between the paraboloids \left\{ \begin{align*} z &= x^2+y^2 \\ z &= 4-x^2-y^2 \end{align*} \right.

To calculate volume, we use $$f=1$$ $$\shaded{ \rm{Volume}=\iiint_R 1\,dV } \nonumber$$

Draw the region $$z(x,y)$$ in $$xyz$$-space

The equations for the paraboloids are given as $$z(x,y)$$, so make $$\int dz$$ the inner integral.

Find the integrals’ bounds:

1. $$\int dz$$: for a fixed point in the $$xy$$-plane, find the bottom and top surface of your solid as $$z(x,y)$$. The $$z$$-bounds follow as $$z_{min}=x^2+y^2$$ and $$z_{max} = 4-x^2-y^2$$.
2. $$\int dy$$: reduce the equation to two variables by looking at the shadow that the solid casts on the $$xy$$-plane. Shadow in $$xyz$$-space Shadow in $$xy$$-plane
Find the radius of the shadow disk in the $$xy$$-plane \begin{align*} z_\text{bottom} &\lt z_\text{top} \\ \Rightarrow x^2+y^2 &\lt 4-x^2-y^2 \\ \Rightarrow 2x^2+2y^2 &\lt 4\\ \Rightarrow x^2+y^2 &\lt 2 \\ \Rightarrow \text{disk }&\text{of radius }\sqrt{2} \end{align*} Consider all $$(x,y)$$ in the shadow. For a given $$x$$, expressing the bounds of $$y$$ gives $$-\sqrt{2-x^2}\lt y\lt \sqrt{2-x^2}$$
3. $$\int dx$$: Consider all $$x$$ in the shadow. The $$x$$-bounds follow as $$-\sqrt{2}\lt x\lt\sqrt{2}$$

Fill in the bounds of the integral $$\int_{-\sqrt{2}}^{\sqrt{2}} \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} \int_{x^2+y^2}^{4-x^2-y^2}\,dz \,dy\,dx \label{eq:ex1a}$$

Evaluate the inner integral \begin{align*} \int_{x^2+y^2}^{4-x^2-y^2}\,dz &= \Big[z\Big]_{z=x^2+y^2}^{4-x^2-y^2} \\ &= 4-x^2-y^2-(x^2+y^2) \\ &= \underline{4-2x^2-2y^2} \end{align*} \nonumber

Evaluate the remaining integral \begin{align*} &\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}} \underline{4-2x^2-2y^2}\,dy\,dx \end{align*} \nonumber

To evaluate this, switch to cylindrical coordinates as described in the next section.

## In cylindrical coordinates

In cylindrical coordinates

• $$r$$ is the distance from the $$z$$-axis
• $$\theta$$ is the angle with the positive $$x$$-axis counterclockwise.
• $$z$$ is the height above the $$xy$$-plane

Some expressions

• $$r=a$$ is a surface in space: a cylinder.
• setting a value for $$\theta$$ will be a vertical half plane starting at the $$z$$-axis (because $$r$$ by convention is always positive)

To convert to $$xyz$$-coordinates \shaded{ (r,\theta,z) \Rightarrow \left\{ \begin{align*} x &= r\,\cos\theta \\ y &= r\,\sin\theta \\ z &= z \end{align*} \right. } \nonumber

### Volume element

A small piece of surface area $$\Delta S$$ at radius $$r$$

The horizontal side of $$\Delta S$$ is $$\Delta r$$. The vertical side is a piece of circle with angle $$\Delta\theta$$ and radius $$r$$ $$\Longrightarrow$$ the length is $$r.\Delta\theta$$. $$\Delta S \approx \Delta r.(r.\Delta\theta) = r\,\Delta r\,\Delta\theta \nonumber$$

With thickness $$\Delta z$$, this small piece has volume $$\Delta V \approx r\,\Delta r\,\Delta\theta\,\Delta z \nonumber$$

The volume element for cylindrical coordinates follows as $$\shaded{ dV = r\,dr\,d\theta\,dz } \nonumber$$

### Examples

#### One

Continue equation $$\eqref{eq:ex1a}$$ from the previous example with the shadow disk $$x^2+y^2\lt 2$$. Keep the inner integral, but switch to the outside integrals to polar coordinates. Input $$(x,y)$$ in $$xyz$$-space

The bounds change as

• The $$\int dz$$-bounds change based on $$x^2+y^2=r^2 \Longrightarrow$$ the upper bound $$4-x^2-y^2=-4-r^2$$.
• The $$\int dr$$ bounds cover the radius of the circle $$0\lt r\lt\sqrt{2}$$.
• The $$\int d\theta$$ bounds cover the angles of the circle $$0\lt \theta\lt 2\pi$$.

The tripple integral in cylindrical coordinates becomes $$\int_0^{2\pi}\int_0^{\sqrt{2}}\int_{r^2}^{4-r^2} \,dz\,r\,dr\,d\theta \nonumber$$ Now, the evaluation is much easier.

## Applications

### Find the mass of a solid

A mass with density $$\delta=\frac{\Delta m}{\Delta v}$$. The mass element $$dm = \delta\,dV \nonumber$$ The mass of a solid $$\shaded{ \rm{mass}=\iiint_R\delta\,dV } \nonumber$$

### Find the average value

The average value of $$f(x,y,z)$$ in $$R$$ $$\shaded{ \bar f=\frac{1}{\rm{Volume}(R)}\iiint_R f\,dV } \nonumber$$

or, the weighted average $$\shaded{ \bar f=\frac{1}{\rm{mass}}\iiint_R f\,\delta\,dV } \nonumber$$

or, center of mass $$(\bar x, \bar y, \bar z)$$, where \shaded{ \begin{align*} \bar x &= \frac{1}{\rm{mass}}\iiint_R x\,\delta\,dV \\ \bar y &= \frac{1}{\rm{mass}}\iiint_R y\,\delta\,dV \\ \bar z &= \frac{1}{\rm{mass}}\iiint_R z\,\delta\,dV \end{align*} } \nonumber

### Find the moment of inertia

The moment of inertia of a solid in respect to an axis $$\iiint_R (\text{distance to axis})^2\,\delta\,dV \nonumber$$

The moment of inertia around the $$x,y,z$$-axis is \shaded{ \begin{align*} I_x &= \iiint_R (y^2+z^2)\,\delta\,dV \\ I_y &= \iiint_R (x^2+z^2)\,\delta\,dV \\ I_z &= \iiint_R (x^2+y^2)\,\delta\,dV \end{align*} } \nonumber Note: rotating around the $$z$$-axis corresponds to rotation around the origin in the plane that we saw in double integrals. The distance to the $$z$$-axis is $$r$$.

### Examples

#### One

Find $$I_z$$ of the solid cone between \left\{ \begin{align*} z &= ar \\ z &= b \end{align*} \right.

The equation $$z=ar$$ implies that $$z$$ is proportional with $$r$$, so a line with slope $$a$$ around the $$z$$-axis. The equation $$z=b$$ represents a plane parallel to the $$xy$$-plane. $$xyz$$-space

Use the equation for the “moment of initia” $$I_z$$ with $$\delta=1$$.

$$I_z = \iiint_R (x^2+y^2)\,dV \nonumber$$

Use cylindrical coordinates, so $$dV=\underline{r\,dr\,d\theta\,dz}$$. \begin{align*} I_z &= \iiint_R (x^2+y^2)\,\underline{r\,dr\,d\theta\,dz} \end{align*}

For the bounds:

• For $$\int dr$$: with $$z$$ constant, we’re slicing the cone in disks of radius $$\frac{z}{a}$$.
• For $$\int d\theta$$, the bounds cover the full circle $$0\lt\theta\lt 2\pi$$.
• For $$\int dz$$ the bounds are limited by the bottom of the cone and the plane at $$z=b$$

Place the bounds on the triple integral and compute \begin{align*} I_z &= \int_0^b \int_0^{2\pi} \int_0^{z/a} r^2\,r\,dr\,d\theta\,dz \\ &= \int_0^b \int_0^{2\pi} \Big[\frac{r^4}{4}\Big]_0^{z/a} \,d\theta\,dz = \frac{1}{4a^4}\int_0^b \int_0^{2\pi}z^4\,d\theta\,dz \\ &= \frac{1}{4a^4}\int_0^b \Big[z^4\theta\Big]_0^{2\pi}\,dz = \frac{2\pi}{4a^4}\int_0^b z^4\,dz \\ &= \frac{2\pi}{4a^4}\Big[\frac{z^5}{5}\Big]_0^b = \frac{\pi}{2a^4}\frac{b^5}{5} = \frac{\pi b^5}{10a^4} \end{align*}

#### Two

Setup a triple integral for the volume inside a unit sphere centered at the origin, and above the plane $$z\gt 1-y$$. $$\left\{ \begin{array}{c} x^2 + y^2 + z^2 \lt 1 \\ z \gt 1-y \end{array} \right. \nonumber$$

The plane $$z\gt 1-y$$, is independent of $$x$$, therefore parallel to the $$x$$-axis. At $$y=0$$, the $$z=1$$. The slope is $$1$$. $$xyz$$-space

The equations for the surfaces are given as z(x,y), so make $$\int dz$$ the inner integral. $$\iint\int_{1+y}^{\sqrt{1-x^2+y^2}} dz\,dx\,dy \nonumber$$

Projected on the on the $$yz$$- and $$xy$$-plane $$z(x,y)$$ on the $$yz$$-plane $$x(y)$$ on the $$xy$$-plane

For the bounds:

• For $$\int dz$$: for a given $$(x,y)$$, $$z_{min}$$ is set by the plane $$z=1+y$$, and $$z_{max}$$ is determined by the sphere $$z=\sqrt{1-x^2+y^2}$$.
• For $$\int dx$$: looked from above, in the $$xy$$-plane, the shadow of the solid is an ellipse. For a given $$y$$, the bounds $$x_{min}$$ and $$x_{max}$$ are when the plane intersects the sphere. \require{cancel} \begin{align*} 1-y &= \sqrt{1-x^2-y^2} \\ \Rightarrow (1-y)^2 &= 1-x^2-y^2 \\ \Rightarrow \bcancel{1}-2y+y^2 &= \bcancel{1}-x^2-y^2 \\ \Rightarrow x^2 &= 2y+2y^2 \\ \Rightarrow x &= \pm\sqrt{2y+2y^2} \end{align*} \nonumber
• For $$\int dy$$, the bounds are $$0$$ and $$1$$.

Place the bounds on the triple integral $$\int_0^1 \int_{-\sqrt{2y+2y^2}}^{\sqrt{2y+2y^2}} \int_{1-y}^{\sqrt{1-x^2-y^2}} dz\,dx\,dy \nonumber$$

## In spherical Coordinates

We use spherical coordinates where there is a lot of symmetry. Best when the solid is centered on the $$z$$-axis.

For spherical coordinates: imagine a vertical half plane through the origin and a point, the $$rz$$-plane.

Spherical coordinates are like “polar coordinates” in this $$rz$$-plane

• $$\rho$$ (rho) is the distance from the origin.
• $$\phi$$ (phi) is the angle form the positive $$z$$-axis, $$0\leq\phi\leq\pi$$.
• $$\theta$$ (theta) is the angle counterclockwise from the positive $$x$$-axis.

Converting spherical $$(\rho,\phi,\theta)$$ $$\to$$ cylindrical coordinates $$(r,\theta,z)$$ \shaded{ (\rho,\phi,\theta) \Rightarrow \left\{ \begin{align*} r &= \rho\sin\phi \\ \theta &= \theta \\ z &= \rho\cos\phi \end{align*} \right. } \nonumber Converting spherical $$(\rho,\phi,\theta)$$ $$\to$$ Cartesian coordinates $$(x,y,z)$$ \shaded{ (\rho,\phi,\theta) \Rightarrow \left\{ \begin{align*} x &= r\cos\theta = \rho\sin\phi\cos\theta \\ y &= r\sin\theta = \rho\sin\phi\sin\theta \\ z &= \rho\cos\phi \end{align*} \right. } \nonumber Converting Cartesian $$(x,y,z)$$ $$\to$$ spherical coordinates $$(\rho,\phi,\theta)$$ \shaded{ (x,y,z) \Rightarrow \left\{ \begin{align*} \rho &= \sqrt{r^2+z^2} = \sqrt{x^2+y^2+z^2} \\ \phi &= \rm{atan}\,\frac{\sqrt{x^2+y^2}}{z} \\ \theta &= \rm{atan}\,\frac{y}{x}\\ \end{align*} \right. } \nonumber

Some expressions

• $$\rho=a$$: a sphere of radius $$a$$, centered at the origin.
• $$\phi=\frac{\pi}{4}$$: a cone. In cylindrical that would $$z=r$$.
• $$\phi=\frac{\pi}{2}$$: the $$xy$$-pane.

### Volume element

To find the volume element $$\Delta V\approx\Delta\rho\,\Delta\phi\,\Delta\theta$$, start with the surface area on a sphere with radius $$a$$. If it is small enough it looks like a rectangle. So, what are the sides? Surface area $$\Delta S$$ Closeup of surface area $$\Delta S$$

Start with a surface area element $$\Delta S$$

• To move north-south: you always have to traverse half a circle $$\Longrightarrow$$ the vertical side of $$\Delta S$$ is a piece of circle with radius $$a$$ $$\Rightarrow$$ the length is $$a.\Delta\phi$$.
• To move east-west: your distance depends on your distance to the $$z$$-axis, called $$r$$ $$\Longrightarrow$$ the horizontal side of $$\Delta S$$ is a piece of circle of radius $$r=a\sin\phi$$. That makes its length $$a.\sin\phi.\Delta\theta$$.

The surface $$\Delta S$$ of this small piece of sphere follows \begin{align*} \Delta S &\approx (a.\sin\phi.\Delta\theta) . (a.\Delta\phi) \\ &\approx a^2\,\sin\phi\,\Delta\phi\,\Delta\theta \end{align*} \nonumber

Moving this area from the surface at radius $$a$$ to a distance $$\rho$$ from the origin $$\Delta S’ \approx \rho^2\,\sin\phi\,\Delta\phi\,\Delta\theta \nonumber$$

Give area $$\Delta S’$$ a thickness $$\Delta\rho$$ Closeup of $$\Delta V$$

This small piece has volume $$\Delta V$$ \begin{align*} \Delta V &\approx \Delta\rho.\Delta S’ \\ &\approx \Delta\rho\,(\rho^2\,\sin\phi.\Delta\phi.\Delta\theta) \end{align*} \nonumber

The volume element for spherical coordinates follows as $$\shaded{ dV = \rho^2\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta } \nonumber$$

### Examples

#### One

Redo a previous example using spherical coordinates: Setup a triple integral for the volume inside $$\left\{ \begin{array}{c} z \gt 1-y \\ x^2 + y^2 + z^2 \lt 1 \end{array} \right. \nonumber$$

To make it easier, we rotate it so that the plane is horizontal, and there is symmetry around the $$z$$-axis $$\left\{ \begin{array}{c} z \gt \frac{1}{\sqrt{2}} \\ x^2 + y^2 + z^2 &\lt 1 \end{array} \right. \nonumber$$

Picture Sphere cap in $$xyz$$-space

Setup triple integral for volume in spherical coordinates $$\iiint_R 1\overbrace{\,\rho^2\,\sin\phi\,d\rho\,d\phi\,d\theta}^{dV} \nonumber$$

For the bounds:

• For $$\int d\rho$$: angles $$\phi$$ and $$\theta$$ are fixed, and $$\rho$$ is the integration variable. It is like shooting a ray from the origin, and seeing where it enters and leaves the solid. We exit the solid cap, when $$\rho=1$$. We enter it when it hits the plane at $$z =\frac{1}{\sqrt{2}}$$. In spherical coordinates that is \begin{align*} z &=\frac{1}{\sqrt{2}} & (z =\rho\cos\phi)\\ \Rightarrow \rho\cos\phi &= \frac{1}{\sqrt{2}} \\ \Rightarrow \rho &= \frac{1}{\sqrt{2}\cos\phi} \end{align*}
• For $$\int d\phi$$: sketch the $$rz$$-plane Sphere cap in $$rz$$-plane
The minimum is just the north pole at $$\phi=0$$. The maximum is where it hits the boundary between the sphere and the plane. \begin{align*} \left. \begin{array}{lc} \rm{sphere:} & \rho = 1 \\ \rm{plane:} & \rho\cos\phi = \frac{1}{\sqrt{2}} \end{array} \right\} \Rightarrow \cos\phi &= \frac{1}{\sqrt{2}} \\ \Rightarrow \phi &= \frac{\pi}{4} \end{align*}
• For $$\int d\theta$$: angle $$\theta$$ goes all around $$\Longrightarrow$$ $$0\lt\theta\lt 2\pi$$

Place the bounds on the triple integral $$\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_{\sec\phi/\sqrt{2}}^1 \,\rho^2\,\sin\phi\,d\rho\,d\phi\,d\theta \nonumber$$

This is quite evil, but eventually evaluates to $$\rm{Volume} = \frac{2\pi}{3}-\frac{5\pi}{6\sqrt{2}} \nonumber$$

### Gravitational attraction

#### In Cartesian coordinates

Express the gravitational force exerted by solid $$M$$ at $$(x,y,z)$$ on a mass $$m$$ at the origin.

With $$G$$ being the gravitational constant, physics tells us

$$|\vec F| = \frac{G.m_1.m_2}{\rm{distance}^2} \nonumber$$

For a tiny mass $$\Delta M$$, substitute \left\{ \begin{align*} m_1 &= m \\ m_2 &\approx \Delta M \\ \rm{distance} &= \rho \end{align*} \right.

Combine this with the direction of this force (unit vector) \left. \begin{align*} |\vec {F_\Delta}| &\approx \frac{G.m.\Delta M}{\rho^2} \\ \text{dir}\left(\vec F_\Delta\right) &= \frac{\left\langle x,y,z \right\rangle}{\rho} \nonumber \end{align*} \right\} \Rightarrow \vec F_\Delta \approx \frac{G.m}{\rho^3}\left\langle x,y,z \right\rangle\,\Delta M

Let the density be $$\delta$$, and substitute $$\Delta M=\delta\,\Delta V$$ $$\vec F_\Delta \approx Gm\frac{\left\langle x,y,z \right\rangle}{\rho^3}\,\delta\,\Delta V \nonumber$$

Sum all $$\Delta V$$, and take the limit for $$\Delta V\to 0$$ gives the gravitational force $$\shaded{ \vec F = Gm\iiint \frac{\left\langle x,y,z \right\rangle}{\rho^3}\,\delta\,dV } \nonumber$$ Here integrating a vector quantity, just means: integrate component by component to get each component of the force. E.g. for the force in direction of the $$x$$-axis $$F_\color{red}x=Gm\iiint \frac{\color{red}x}{\rho^3}\,\delta\,dV \nonumber$$

The $$\rho^3$$ wrecks havoc with rectangular coordinates, and it will cause a factor like $$(x^2+y^2+z^2)^{3/2}$$. On the other hand, with spherical coordinates it might just disappear as we will see in the following section.

#### In spherical coordinates

Redo the previous example in spherical coordinates: Place the solid so that $$z$$-axis is an axis of symmetry. Symmetrical on $$z$$-axis

Because of the symmetric in the $$z$$-axis, the force $$\vec F$$ will only have a $$z$$-component \left\{ \begin{align*} \vec F &= \left\langle 0,0,F_z\right\rangle \\ F_z &= Gm\iiint \frac{z}{\rho^3}\delta\,dV \end{align*} \right.

Using spherical coordinates, $$z=\rho\cos\phi$$ and $$dV=\rho^2\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta$$ \begin{align*} F_z &= Gm\iiint \frac{\bcancel{\rho}\cos\phi}{\bcancel{\rho^3}} \delta\,\underbrace{\bcancel{\rho^2}\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta}_{dV} \\ &= Gm\iiint \delta\,\cos\phi\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta \end{align*} \nonumber

### Proof Newton’s theorem

Continuing from the gravitational attration example: proof Newton’s theorem.

The gravitational attraction of a spherical planet with uniform density is equal to that of a point mass (with the same total mass) at its center.

Setup: the sphere has radius $$a$$ and is centered on the positive $$z$$-axis, tangent to the $$xy$$-plane at the origin $$M$$ and $$m$$ in $$xyz$$-space

Let $$M$$ be the mass of the planet, and $$m$$ be the test mass at the origin $$F_z = Gm\iiint \delta\,\cos\phi\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta \nonumber$$

For the bounds

• For the inner integral $$\int d\rho$$, sketch a slice in the $$rz$$-plane to find the upper bound for $$\rho$$ using the right triangle, $$\rho_{max}=2a\cos\phi$$ $$M$$ and $$m$$ in $$rz$$-plane
• For the middle integral $$\int d\phi$$, the bounds are $$0$$ to $$\pi/2$$.
• For the outer integral $$\int d\theta$$, the bounds are $$0$$ to $$2\pi$$.

Enter the bounds and evaluate $$F_z = Gm\int_0^{2\pi} \int_0^{\pi/2} \underline{\int_0^{2a\cos\phi} \delta\,\cos\phi\,\sin\phi\,\,d\rho}\,\,d\phi\,\,d\theta \nonumber$$

The inner integral \begin{align*} \int_0^{2a\cos\phi} \delta\,\cos\phi\,\sin\phi\,\,d\rho &= \delta\cos\phi\,\sin\phi\,\Big[\rho\Big]_{\rho=0}^{2a\cos\phi} \\ &= \delta\cos\phi\sin\phi\,2a\cos\phi \\ &= \underline{2a\delta\cos^2\phi\sin\phi} \end{align*}

For the middle integral, substitute $$u=\cos\phi \Rightarrow du=-\sin\phi\,d\phi$$ \begin{align*} \int_0^{\pi/2}\underline{2a\delta\,\cos^2\phi\,\sin\phi}\,\,d\phi &= \int_0^{\pi/2}-2a\,\delta\,\overbrace{\cos^2\phi}^{u^2}\,(\overbrace{-\sin\phi\,\,d\phi}^{du}) \\ &= -2a\delta \int u^2\,du = -2a\delta \Big[\frac{u^3}{3}\Big]_{\cos0}^{u=\cos\pi/2} \\ &= -2a\delta\left(0-\frac{1}{3}\right) = \underline{\frac{2}{3}a\delta} \end{align*}

The outer integral \begin{align*} F_z &= Gm\int_0^{2\pi}\underline{\frac{2}{3}a\delta}\,d\theta = Gm\,\frac{2}{3}a\delta\Big[\phi\Big]_{0}^{\phi=2\pi}\\ &=\frac{2}{3}a\delta\,Gm\,2\pi = \underline{\frac{4}{3}\pi a\delta}\,Gm \end{align*}

The volume of a spherical planet with radius $$a$$ is $$\frac{4}{3}\pi a^3$$ $$\Longrightarrow$$ the mass of this planet with uniform density $$\delta$$ $$=$$ volume $$\times$$ density. \begin{align*} M &= \frac{4}{3}\pi a^3 \delta \Longleftrightarrow \underline{\frac{4}{3}\pi a\delta=\frac{M}{a^2}} \end{align*} \nonumber

Substituting $$\frac{M}{a^2}$$ in the outer integral proves Newton’s theorem $$\shaded{ F_z = \frac{GmM}{a^2} } \nonumber$$

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