2 Surface and volume integrals

2.1 Surfaces

A 2-d surface embedded in \(\Real^3\) can be defined parametrically as a map \(\xb(u,v)\) from \(\Real^2\to\Real^3\), for \(u\in[u_0,u_1]\) and \(v\in[v_0,v_1]\).

For each value of \(v\), the position \(\xb(u,v)\) describes a curve on the surface with tangent vector \(\displaystyle\ddy{\xb}{u}\).

Similarly, for each value of \(u\), we get a curve with tangent vector \(\displaystyle\ddy{\xb}{v}\).

If \((u,v)\) is a well-defined parametrisation, then their tangent vectors cannot be collinear. So the cross product of these tangent vectors allows us to define the unit normal to the surface: \[ \hat{\nb} = \frac{\displaystyle\ddy{\xb}{u}\times\ddy{\xb}{v}}{\displaystyle\left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right|}. \]

The sign of \(\hat{\nb}\) depends on the ordering of \((u,v)\). A surface together with a specified (consistent) choice of \(\hat{\nb}\) is called an oriented surface.

Not all surfaces are orientable, meaning that a consistent choice of \(\hat{\nb}\) exists. An example of a non-orientable surface would be a Möbius strip.

Example 2.1: Find the outward unit normal for the sphere \(x^2+y^2+z^2=a^2\).

We know from our definition of spherical coordinates that the (outward) unit normal is given by \(\hat{\nb} = \eb_r\) (one of the basis vectors in spherical coordinates).

[You could recover this from the position vector \[ \xb(\theta,\phi)=a\sin\theta\cos\phi\eb_1 + a\sin\theta\sin\phi\eb_2 + a\cos\theta\eb_3, \] by taking \((u,v)=(\theta,\phi)\) and computing \[ \ddy{\xb}{\theta}\times\ddy{\xb}{\phi} = (h_\theta\eb_\theta)\times(h_\phi\eb_\phi) = h_\theta h_\phi\eb_r = a^2\sin\theta\eb_r. \] Divide by the magnitude \(a^2\sin\theta\) to get \(\eb_r\).]

Example 2.2: Find the outward unit normal for the torus \(\xb(u,v)=(R+\cos{v})\cos{u}\eb_1 + (R+\cos{v})\sin{u}\eb_2 + \sin{v}\eb_3\), where \(R>1\) and \(u,v\in[0,2\pi]\).

The tangent vectors are \[ \ddy{\xb}{u} = -(R+\cos{v})\sin{u}\eb_1 + (R+\cos{v})\cos{u}\eb_2, \quad \ddy{\xb}{v}=-\sin{v}\cos{u}\eb_1 - \sin{v}\sin{u}\eb_2 + \cos{v}\eb_3, \] so \[\begin{align*} \ddy{\xb}{u}\times\ddy{\xb}{v} &= \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ -(R+\cos{v})\sin{u} & (R+\cos{v})\cos{u} & 0\\ -\sin{v}\cos{u} & -\sin{v}\sin{u} & \cos{v} \end{vmatrix}\\ &= (R+\cos{v})\cos{v}\cos{u}\eb_1 + (R+\cos{v})\cos{v}\sin{u}\eb_2 + (R+\cos{v})\sin{v}\eb_3\\ &= (R+\cos{v})\Big(\cos{v}\cos{u}\eb_1 + \cos{v}\sin{u}\eb_2 + \sin{v}\eb_3\Big). \end{align*}\] This has magnitude \[ \left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right| = |R+\cos{v}|\sqrt{\cos^2{v}\cos^2{u} + \cos^2{v}\sin^2{u} + \sin^2{v}} = R+\cos{v}, \] where the absolute value sign can be removed since \(R>1\). So the unit normal is \[ \hat{\nb} = \cos{v}\cos{u}\eb_1 + \cos{v}\sin{u}\eb_2 + \sin{v}\eb_3. \]

[This is a doughnut-shaped surface as in the following picture.]

A surface is called smooth if \(\hat{\nb}\) varies continuously over the surface (e.g. a sphere). It is called piecewise smooth if it can be divided into finitely many portions each of which is smooth (e.g. a cube).

Example 2.3: Sketch the surface \(\xb(r,\theta)=r\cos{\theta}\eb_1 + r\sin{\theta}\eb_2 + r\eb_3\) for \(r\in[0,\infty)\) and \(\theta\in[0,2\pi]\) and determine whether or not it is smooth.

To sketch the surface, note that we have \(x=r\cos{\theta}\), \(y=r\sin{\theta}\) and \(z=r\), so \[ x^2 + y^2 = r^2 = z^2 \quad \implies z=\sqrt{x^2+y^2}. \] This is a cone with its point at \(\xb=\bfzero\).

The surface is not smooth because \(\hat{\nb}\) is not defined at \(\xb=\bfzero\). To show this algebraically, note that the tangent vectors are \[ \ddy{\xb}{r} = \cos{\theta}\eb_1 + \sin{\theta}\eb_2 + \eb_3, \quad \ddy{\xb}{\theta}=-r\sin{\theta}\eb_1 + r\cos{\theta}\eb_2 \] so \[ \ddy{\xb}{r}\times\ddy{\xb}{\theta} = \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ \cos\theta & \sin\theta & 1\\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix}= -r\cos{\theta}\eb_1 - r\sin{\theta}\eb_2 + r\eb_3 = -x\eb_1 -y\eb_2 +z\eb_3. \] So indeed this vanishes at \(\xb=\bfzero\).

A surface can be either open if it is bounded by a curve (e.g. the cone, or a hemisphere), or closed if it has no boundary (e.g. the sphere or torus). The boundary of an open surface \(S\) is a curve (or union of curves), often written \(\partial S\).

A simple type of surface is an explicit surface given by the height of a two-dimensional function, \(z=h(x,y)\).

Example 2.4: Find the upward unit normal to the surface \(z=h(x,y)\).

We can parametrize the surface just with \(x\) and \(y\) so \(\xb(x,y)=x\eb_1 + y\eb_2 + h(x,y)\eb_3\).

[Example 2.3 is an example of this type of surface with \(h(x,y)=\sqrt{x^2+y^2}\).]

The tangent vectors are then \[ \ddy{\xb}{x} = \eb_1 + \ddy{h}{x}\eb_3, \quad \ddy{\xb}{y} = \eb_2 + \ddy{h}{y}\eb_3, \] so \[ \ddy{\xb}{x}\times\ddy{\xb}{y} = \begin{vmatrix} \eb_1 & \eb_2 & \eb_3\\ 1 & 0 & \displaystyle\ddy{h}{x}\\ 0 & 1 & \displaystyle\ddy{h}{y} \end{vmatrix} = -\ddy{h}{x}\eb_1 - \ddy{h}{y}\eb_2 + \eb_3. \] This is clearly pointing upward, so the required unit normal is \[ \hat{\nb} = \frac{\displaystyle-\ddy{h}{x}\eb_1 - \ddy{h}{y}\eb_2 + \eb_3}{\displaystyle\sqrt{1 + \left(\ddy{h}{x}\right)^2 + \left(\ddy{h}{y}\right)^2}}. \]

2.2 Surface integrals of scalar fields

If \(f(\xb)=f(x,y,z)\) is a scalar field and \(S\) is a surface with parametrisation \(\xb(u,v)\), then we define the surface integral of \(f\) over \(S\) to be \[ \int_S f\,dS = \int_Uf(\xb(u,v))\left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right|\,du\,dv, \] where \(U\) is the preimage of \(S\) in the \((u,v)\) plane.

To motivate this formula, note that the area element \(\displaystyle\left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right|\,du\,dv\) is the area on the surface in \(\Real^3\) corresponding to an infinitesimal rectangle with sides \((du, dv)\) in the \((u,v)\) plane.

Example 2.5: Find the area element for the plane \(z=ax+b\) when parametrised by \((x,y)\).

The parametrisation gives \(\xb(x,y)=x\eb_1 + y\eb_2 + (ax+b)\eb_3\). This is an “explicit surface” of the form \(z=h(x,y)\), so from Example 2.4 we have \[ \left|\ddy{\xb}{x}\times\ddy{\xb}{y}\right| = \sqrt{1 + \left(\ddy{h}{x}\right)^2 + \left(\ddy{h}{y}\right)^2} = \sqrt{1 + a^2}. \] So the area element is \(\sqrt{1+a^2}\,dx\,dy\).

[Interpretation: when \(a=0\), the plane is flat and areas on the surface are identical to those in the \((x,y)\) plane. When \(a\neq 0\), areas on the surface are greater than those in the \((x,y)\) plane, due to the tilt of the surface.]

Proposition 2.1

If \((u,v)\) are (two of a set of) orthogonal curvilinear coordinates, then \[ \left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right| = h_uh_v. \]

Proof. Recall that \(\displaystyle\ddy{\xb}{u}=h_u\eb_u\) and \(\displaystyle\ddy{\xb}{v}=h_v\eb_v\). Since \(h_u,h_v \geq 0\), we have that \[ \left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right| = h_uh_v|\eb_u\times\eb_v| = h_uh_v|\pm\eb_w| = h_uh_v. \]

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The surface area of \(S\) is computed by taking \(f(\xb)\equiv 1\), giving \(\displaystyle|S|=\int_S\,dS\).

Example 2.6: Use spherical coordinates to find the surface area of the sphere \(x^2+y^2+z^2=a^2\).

The sphere radius \(a\) is given by \(\xb(\theta,\phi) = a\sin\theta\cos\phi\eb_1 + a\sin\theta\sin\phi\eb_2 + a\cos\theta\eb_3\), and we know that the scale factors are \(h_\theta=a\) and \(h_\phi=a\sin\theta\). So from Proposition 2.1 we see that \[ \left|\ddy{\xb}{\theta}\times\ddy{\xb}{\phi}\right|=h_\theta h_\phi = a^2\sin\theta. \] The surface area is then \[ |S| = \int_S\,dS= \int_0^{2\pi}\left(\int_0^\pi a^2\sin\theta\,d\theta\right)\,d\phi= a^2\int_0^{2\pi}\Big[-\cos\theta\Big]_0^{\pi}\,d\phi = 2a^2\int_0^{2\pi}\,d\phi = 4\pi a^2. \]

Example 2.7: Find the surface area of the torus from Example 2.2.

Recall that our parametrisation \(\xb(u,v)\) had \(u,v\in[0,2\pi]\) and \(\displaystyle\left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right|=R+\cos{v}\), so \[ |S| = \int_S\,dS = \int_0^{2\pi}\left(\int_0^{2\pi}(R + \cos{v})\,du\right)\,dv = \int_0^{2\pi}2\pi(R+\cos{v})\,dv = 2\pi\Big[Rv + \sin{v}\Big]_0^{2\pi} = 4\pi^2R. \]

Notice that the value of a scalar surface integral does not depend on the sign of \(\hat{\nb}\), so is independent of the choice of orientation of a surface (or indeed whether it is orientable at all).

Proposition 2.2

The surface integral of a scalar field is independent of the choice of parametrisation.

Proof. Suppose \(\xb(u,v)\) and \(\xb(\mu,\nu)\) are two regular parametrisations of the same surface. By the chain rule, \[ \ddy{\xb}{u} = \ddy{\xb}{\mu}\ddy{\mu}{u} + \ddy{\xb}{\nu}\ddy{\nu}{u}, \qquad \ddy{\xb}{v} = \ddy{\xb}{\mu}\ddy{\mu}{v} + \ddy{\xb}{\nu}\ddy{\nu}{v}. \] So \[\begin{align*} \ddy{\xb}{u}\times\ddy{\xb}{v} &= \left(\ddy{\xb}{\mu}\ddy{\mu}{u} + \ddy{\xb}{\nu}\ddy{\nu}{u}\right)\times\left(\ddy{\xb}{\mu}\ddy{\mu}{v} + \ddy{\xb}{\nu}\ddy{\nu}{v} \right)\\ &= \left(\ddy{\xb}{\mu}\ddy{\mu}{u} \times \ddy{\xb}{\nu}\ddy{\nu}{v}\right) + \left(\ddy{\xb}{\nu}\ddy{\nu}{u} \times \ddy{\xb}{\mu}\ddy{\mu}{v}\right)\\ &= \left(\ddy{\mu}{u}\ddy{\nu}{v} - \ddy{\mu}{v}\ddy{\nu}{u}\right)\ddy{\xb}{\mu}\times\ddy{\xb}{\nu}. \end{align*}\] This means \[ \left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right| = |J|\left|\ddy{\xb}{\mu}\times\ddy{\xb}{\nu}\right|, \] where \(\displaystyle J= \ddy{\mu}{u}\ddy{\nu}{v} - \ddy{\mu}{v}\ddy{\nu}{u}\) is the Jacobian from Calculus I for the change of variables from \((\mu, \nu)\) to \((u,v)\). Since \(|J|\,du\,dv=d\mu\,d\nu\), we obtain \[ \int_Uf(\xb(u,v))\left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right|\,du\,dv = \int_{U'}f(\xb(\mu,\nu))\left|\ddy{\xb}{\mu}\times\ddy{\xb}{\nu}\right|\,d\mu\,d\nu, \] where \(U'\) is the corresponding region of the \((\mu,\nu)\) plane.

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If \(S\subset\Real^2\) is a flat surface in (say) the \((x,y)\)-plane, then \(\hat{\nb}\) only has a \(z\)-component, so \[ \left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right| = \left|\ddy{x}{u}\ddy{y}{v} - \ddy{x}{v}\ddy{y}{u}\right| = |J|, \] which is just the Jacobian for a change of coordinates from \((x,y)\) to \((u,v)\) in the plane.

2.3 Surface integrals of vector fields

If \(\fb:\Real^3\to\Real^3\) is a vector field and \(S\) is an oriented surface with unit normal \(\hat{\nb}\), we define the surface integral (or flux) of \(\fb\) through \(S\) as \[ \int_S\fb\cdot\hat{\nb}\,dS. \]

Sometimes you will see the shorthand notation \(\dS = \hat{\nb}\,dS\) for the “vector area element”.

A non-zero flux requires a component of \(\fb\) pointing “through” the surface.

For given \(|\fb|\), the flux is largest if \(\fb\) is exactly perpendicular to the surface, but the flux also depends on the magnitude \(|\fb|\).

Again, we evaluate the integral using a specific parametrisation \(\xb(u,v)\), so \[\begin{align*} \int_S\fb\cdot\hat{\nb}\,dS &= \int_U \fb(\xb(u,v))\cdot\hat{\nb}\left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right|\,du\,dv\\ &= \int_U \fb(\xb(u,v))\cdot\left(\ddy{\xb}{u}\times\ddy{\xb}{v}\right)\,du\,dv, \end{align*}\] where \(U\) is the preimage of \(S\) in the \((u,v)\) plane.

Unlike a scalar surface integral, a flux integral depends on the sign of \(\hat{\nb}\), i.e. on the orientation of the surface. You need to check that your chosen parametrisation \(\xb(u,v)\) gives the correct sign of \(\hat{\nb}=\displaystyle\ddy{\xb}{u}\times\ddy{\xb}{v}\). If not, interchange \(u\) and \(v\).

Example 2.8: Calculate the outward flux of \(\fb(\xb)=\cos^7\theta\,\eb_r\) [in spherical coordinates] through the hemisphere \(r=a\), \(z\geq 0\).

Parametrising the sphere in spherical coordinates as \(\xb(\theta,\phi)\), recall that \(\hat{\nb}=\eb_r\) and \(\displaystyle\left|\ddy{\xb}{\theta}\times\ddy{\xb}{\phi}\right|=h_\theta h_\phi = a^2\sin\theta\).

Also note that the hemisphere is given by the half-range \(\theta\in[0,\pi/2]\), with \(\phi\in[0,2\pi]\) as usual.

So \[\begin{align*} \int_S\fb\cdot\hat{\nb}\,dS &= \int_0^{2\pi}\left(\int_0^{\pi/2}\fb\cdot\hat{\nb}\left|\ddy{\xb}{\theta}\times\ddy{\xb}{\phi}\right|\,d\theta\right)\,d\phi\\ &= \int_0^{2\pi}\left(\int_0^{\pi/2}(\cos^7\theta\,\eb_r)\cdot(\eb_r)a^2\sin\theta\,d\theta\right)\,d\phi\\ &= a^2\int_0^{2\pi}\left(\int_0^{\pi/2}\cos^7\theta\sin\theta\,d\theta\right)\,d\phi\\ &= a^2\int_0^{2\pi}\left[-\frac{1}{8}\cos^8\theta\right]_0^{\pi/2}\,d\phi = \frac{a^2}{8}\int_0^{2\pi}\,d\phi = \frac{\pi a^2}{4}. \end{align*}\]

[This is a model for the Sun’s surface magnetic field at the minimum of the 11-year sunspot cycle (when there are no sunspots). What we have computed would be the magnetic flux through the northern hemisphere (in appropriate units). By contrast, the Earth’s magnetic field looks more like \(\fb=\cos\theta\eb_r\).]

If the surface \(S\) is closed, this is often emphasised by writing \[ \int_S\fb\cdot\hat{\nb}\,dS = \oint_S\fb\cdot\hat{\nb}\,dS. \]

Unlike the circulation around a closed curve, there is no special name for the flux through a closed surface.

Example 2.9: Repeat Example 2.8 for the full sphere.

Now \(\theta\in[0,\pi]\). But recall that \(z=a\cos\theta\), so we have \(\displaystyle\fb=\frac{z^7}{a^7}\eb_r\).

This vector field is antisymmetric under reflection in the plane \(z=0\), as is the surface itself, so the southern hemisphere flux will be the negative of the northern hemisphere flux. So they will cancel giving \[ \oint_S\fb\cdot\hat{\nb}\,dS = 0. \]

We will also meet vector-valued surface integrals of the form \[ \int_S\fb\,dS \] for some vector field \(\fb:\Real^3\to\Real^3\).

Example 2.10: Evaluate \(\displaystyle\int_S\eb_r\,dS\) where \(S\) is the hemisphere from Example 2.8 and \(\eb_r\) is the spherical basis vector.

It’s important to remember that \(\eb_r\) is not constant over the surface, because its direction varies. So we can’t take it out of the integral even though its magnitude is constant. To deal with this, write it in the Cartesian basis [Tutorial Sheet 1]: \[ \eb_r = \sin\theta\cos\phi\eb_1 + \sin\theta\sin\phi\eb_2 + \cos\theta\eb_3. \] Now the unit vectors are constant and can be taken out of the integral: \[\begin{align*} \int_S\eb_r\,dS &= \eb_1\int_S\sin\theta\cos\phi\,dS + \eb_2\int_S\sin\theta\sin\phi\,dS + \eb_3\int_S\cos\theta\,dS. \end{align*}\] The first two integrals will vanish because the \(\phi\) part is the integral over a whole period of cosine or sine. So \[\begin{align*} \int_S\eb_r\,dS &= \eb_3\int_S\cos\theta\,dS\\ &= \eb_3\int_0^{2\pi}\int_0^{\pi/2}\cos\theta(a^2\sin\theta)\,d\theta\,d\phi\\ &= 2\pi a^2\eb_3\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta\\ &= \pi a^2\eb_3\int_0^{\pi/2}\sin{2\theta}\,d\theta = \pi a^2\eb_3\Big[-\frac12\cos{2\theta}\Big]_0^{\pi/2} = \pi a^2\eb_3. \end{align*}\]

Note that this is not the same as \(\eb_r\displaystyle\int_S\,dS = 2\pi a^2\eb_r\), which differs both in magnitude and direction.

[If we divide our original answer by the surface area \(2\pi a^2\), what we have computed is the “average” of the vector \(\eb_r\) over the hemisphere. Purely from symmetry, you can see that the direction had to come out as \(\eb_3\).]

2.4 Volume integrals

The volume integral of a scalar field \(f:\Real^3\to\Real\) over a 3-d subregion \(V\subset\Real^3\) is \[ \int_Vf\,dV = \int_Uf(\xb(u,v,w))\left|\ddy{\xb}{u}\cdot\left(\ddy{\xb}{v}\times\ddy{\xb}{w}\right)\right|\,du\,dv\,dw, \] where \(U\) is the preimage of \(V\) in \((u,v,w)\) space, for some well-defined coordinates \((u,v,w)\).

The factor \(\displaystyle \left|\ddy{\xb}{u}\cdot\left(\ddy{\xb}{v}\times\ddy{\xb}{w}\right)\right|\,du\,dv\,dw\) is the volume of a parallelepiped in “real space” corresponding to an infinitesimal cuboid \((du,dv,dw)\) in \((u,v,w)\) space.

Proposition 2.3

If \((u,v,w)\) are orthogonal curvilinear coordinates, then \[ \left|\ddy{\xb}{u}\cdot\left(\ddy{\xb}{v}\times\ddy{\xb}{w}\right)\right| = h_uh_vh_w. \]

Proof. Recall that \(\displaystyle\ddy{\xb}{u}=h_u\eb_u\), and similarly for the other two tangent vectors. Since \(h_u,h_v,h_w \geq 0\), \[ \left|\ddy{\xb}{u}\cdot\left(\ddy{\xb}{v}\times\ddy{\xb}{w}\right)\right| = h_uh_vh_w |\eb_u\cdot(\eb_v\times\eb_w)| = h_uh_vh_w|\eb_u\cdot(\pm\eb_u)| = h_uh_vh_w. \]

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The factor \(\displaystyle \left|\ddy{\xb}{u}\cdot\left(\ddy{\xb}{v}\times\ddy{\xb}{w}\right)\right|\) is often called the Jacobian (or Jacobian determinant). To see why, note that the scalar triple product may be written \[ \ddy{\xb}{u}\cdot\left(\ddy{\xb}{v}\times\ddy{\xb}{w}\right) = \begin{vmatrix} \displaystyle\ddy{x}{u} & \displaystyle\ddy{x}{v} & \displaystyle\ddy{x}{w}\\ \displaystyle\ddy{y}{u} & \displaystyle\ddy{y}{v} & \displaystyle\ddy{y}{w}\\ \displaystyle\ddy{z}{u} & \displaystyle\ddy{z}{v} & \displaystyle\ddy{z}{w} \end{vmatrix}. \] Compare this to the 2-d case of a flat surface in the \((x,y)\) plane, where we had \[ \left|\ddy{\xb}{u}\times\ddy{\xb}{v}\right| = \begin{vmatrix} \displaystyle\ddy{x}{u} & \displaystyle\ddy{x}{v}\\ \displaystyle\ddy{y}{u} & \displaystyle\ddy{y}{v} \end{vmatrix}. \]

The volume of \(V\) is computed by taking \(f(\xb)\equiv 1\), giving \(\displaystyle|V|=\int_V\,dV\).

Example 2.11: Find the volume of the sphere \(x^2+y^2+z^2=a^2\).

The sphere is given by spherical coordinates \(\xb(r,\theta,\phi)=r\sin\theta\cos\phi\eb_1 + r\sin\theta\sin\phi\eb_2 + r\cos\theta\eb_3\) for \(r\in[0,a]\) and the usual ranges of \(\theta,\phi\).

By Proposition 2.3 the Jacobian is \(\displaystyle\left|\ddy{\xb}{r}\cdot\left(\ddy{\xb}{\theta}\times\ddy{\xb}{\phi}\right)\right| = h_rh_\theta h_\phi = r^2\sin\theta\).

So the volume is \[\begin{align*} |V| &= \int_0^{2\pi}\left(\int_0^\pi\left(\int_0^a r^2\sin\theta\,dr\right)\,d\theta\right)\,d\phi\\ &= \int_0^{2\pi}\left(\int_0^\pi\sin\theta\left[\frac{r^3}{3}\right]_0^a\,d\theta\right)\,d\phi = \frac{a^3}{3}\int_0^{2\pi}\Big[-\cos\theta\Big]_0^{\pi}\,d\phi = \frac{2a^3}{3}\int_0^{2\pi}\,d\phi = \frac43\pi a^3. \end{align*}\]

In more complicated cases, the main difficulty is finding the correct integration limits. It is essential to draw a diagram!

Example 2.12: Let \(V\) be the region bounded by the planes \(x=0\), \(y=0\), \(z=2\) and the surface \(z=x^2+y^2\) for \(x,y\geq 0\). Calculate \(\displaystyle\int_V x\,dV\).

First we make a sketch to establish the shape of the region. Note that the surface \(z=x^2+y^2\) is a paraboloid.

Clearly this region has cylindrical symmetry, so the integral will be easiest to evaluate in cylindrical coordinates, \(\xb(r,\theta,z) = r\cos\theta\eb_1 + r\sin\theta\eb_2 + z\eb_3\).

By Proposition 2.3 the Jacobian is \(\displaystyle\left|\ddy{\xb}{r}\cdot\left(\ddy{\xb}{\theta}\times\ddy{\xb}{z}\right)\right| = h_r h_\theta h_z = r\).

In our case, the boundaries \(x=0\) and \(y=0\) correspond to \(\theta=0\) and \(\theta=\pi/2\), respectively. The range in \(r\) depends on the height, with maximum given by \(z=x^2+y^2=r^2\), so that \(r(z)\in[0,\sqrt{z}]\). Therefore we have \[\begin{align*} \int_V x\,dV &= \int_0^2\left(\int_0^{\pi/2}\left(\int_0^{\sqrt{z}} x r\,dr\right)\,d\theta\right)\,dz\\ &= \int_0^2\left(\int_0^{\pi/2}\left(\int_0^{\sqrt{z}} r^2\cos\theta\,dr\right)\,d\theta\right)\,dz\\ &= \int_0^2\left(\int_0^{\pi/2}\cos\theta\left[\frac{r^3}{3}\right]_0^{\sqrt{z}}\,d\theta\right)\,dz\\ &= \int_0^2\frac{z^{3/2}}{3}\left(\int_0^{\pi/2}\cos\theta\,d\theta\right)\,dz\\ &= \int_0^2\frac{z^{3/2}}{3}\Big[\sin\theta\Big]_0^{\pi/2}\,dz= \int_0^2\frac{z^{3/2}}{3}\,dz= \frac{2}{15}\Big[z^{5/2}\Big]_0^2 = \frac{2}{15}2^{5/2} = \frac{8\sqrt{2}}{15}. \end{align*}\]

Proposition 2.4

The volume integral of a scalar field is independent of the choice of coordinates.

Proof. This is analogous to Proposition 2.2, using the fact that a 3-d change of variables obeys \[ dx\,dy\,dz = \left|\ddy{\xb}{u}\cdot\left(\ddy{\xb}{v}\times\ddy{\xb}{w}\right)\right|\,du\,dv\,dw. \]

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Occasionally you will come across a vector-valued volume integral, \(\displaystyle\int_V\fb\,dV\).

An example is the centre of mass of a solid body, \[ \overline{\xb} = \frac{\displaystyle\int_V\rho(\xb)\xb\,dV}{\displaystyle\int_V\rho(\xb)\,dV} \] where \(\rho(\xb)\) is the distribution of mass density within the body.