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, 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.


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

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


  • 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 $$ \shaded{ dV=dx\,dy\,dz } \nonumber $$



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
Cylindrical coordinates

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 \(r\)

Cylindrical coordinates
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 $$ \shaded{ dV = r\,dr\,d\theta\,dz } \nonumber $$



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.


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 \(r\).



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.


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 $$ \begin{align*} I_z &= \int_0^b \int_0^{2\pi} \int_0^{z/a} r^2\,r\,dr\,d\theta\,dz \\ &= \int_0^b \int_0^{2\pi} \Big[\frac{r^4}{4}\Big]_0^{z/a} \,d\theta\,dz = \frac{1}{4a^4}\int_0^b \int_0^{2\pi}z^4\,d\theta\,dz \\ &= \frac{1}{4a^4}\int_0^b \Big[z^4\theta\Big]_0^{2\pi}\,dz = \frac{2\pi}{4a^4}\int_0^b z^4\,dz \\ &= \frac{2\pi}{4a^4}\Big[\frac{z^5}{5}\Big]_0^b = \frac{\pi}{2a^4}\frac{b^5}{5} = \frac{\pi b^5}{10a^4} \end{align*} $$


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 \(1\).


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

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 $$ \shaded{ dV = \rho^2\,\sin\phi\,\,d\rho\,\,d\phi\,\,d\theta } \nonumber $$



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 $$ \left\{ \begin{array}{c} z \gt \frac{1}{\sqrt{2}} \\ x^2 + y^2 + z^2 &\lt 1 \end{array} \right. \nonumber $$


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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