After completing this section you will...
Understand how to compute a little bit of surface area
Revisit parameterizing surfaces
Revisit finding vectors normal to a surface
In first-semester calculus, we learned how to compute integrals of the form \(\int_a^b f dx\) along straight (flat) segments \([a,b]\text{.}\) This semester, in the line integral unit, we learned how to change the segment to a curve, which allowed us to compute integrals of the form \(\int_C fds\) along any curve \(C\text{,}\) instead of just along curves (segments) on the \(x\)-axis.
The integral \(\int_a^b dx=b-a\) gives the length of the segment \([a,b]\text{.}\)
The integral \(\int_C ds\) gives the length \(s\) of the curve \(C\text{.}\)
In the double integral unit we learned how to compute double integrals \(\iint_R fdA\) along flat regions \(R\) in the plane. In this unit we will now learn how to change the flat region \(R\) into a curved surface \(S\text{,}\) and then compute integrals of the form \(\iint_S fd\sigma\) along curved surfaces. The differential \(d\sigma\) stands for a little bit of surface area. We already know that \(\iint_R dA\) gives the area of \(R\text{.}\) We'll define \(\iint_S d\sigma\) so that it gives the surface area of \(S\text{.}\)
Exercise12.1.1
Consider the surface \(S\) given by \(z=9-x^2-y^2\) (we've seen this surface many times). A parametrization of this surface is
\begin{equation*}
\vec r(x,y) = (x,y,9-x^2-y^2).
\end{equation*}
(a)
Draw the surface \(S\text{.}\)
(b)
Add to your surface plot the parabolas given by \(\vec r(x,0)\text{,}\) \(\vec r(x,1)\text{,}\) and \(\vec r(x,2)\text{,}\) as well as the parabolas given by \(\vec r(0,y)\text{,}\) \(\vec r(1,y)\text{,}\) and \(\vec r(2,y)\text{.}\)
(c)
Find \(\ds \frac{\partial \vec r}{\partial x}\) and \(\ds\frac{\partial \vec r}{\partial y}\text{.}\)
(d)
At the point \((2,1)\text{,}\) draw both vectors.
(e)
These vectors form the edges of a parallelogram. Add that parallelogram to your picture.
(f)
Show that the area of a parallelogram whose edges are the vectors \(\ds\frac{\partial \vec r}{\partial x}(x,y)\) and \(\ds\frac{\partial \vec r}{\partial y}\) is \(\sqrt{1+4x^2+4y^2}\text{.}\)
HintWhat was one part of the definition/meanings for the cross product.
(g)
Find the area of the parallelogram whose edges are the vectors \(\ds\frac{\partial \vec r}{\partial x}dx\) and \(\ds\frac{\partial \vec r}{\partial y}dy\text{,}\) where \(dx\) and \(dy\) are not yet determined.
In the previous problem, you showed that the area of the parallelogram with edges given by \(\frac{\partial \vec r}{\partial x}dx\) and \(\frac{\partial \vec r}{\partial y}dy\) is
\begin{equation*}
d\sigma =\left |\frac{\partial \vec r}{\partial x} \times \frac{\partial \vec r}{\partial y}\right| dxdy.
\end{equation*}
This little bit of area approximates the area of a tiny patch on the surface. If we add all these areas up, we should obtain the surface area.
Definition12.1.1
Let \(S\) be a surface. Let \(\vec r(u,v)=(x,y,z)\) be a parametrization of the surface, where the bounds on \(u\) and \(v\) form a region \(R\) in the \(uv\) plane. Then the surface area element (representing a little bit of surface) is
\begin{equation*}
d\sigma =\left |\frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v}\right| dudv = \left|\vec r_u\times\vec r_v\right|dudv.
\end{equation*}
The surface integral of a continuous function \(f(x,y,z)\) along the surface \(S\) is
\begin{equation*}
\iint_S f(x,y,z) d\sigma = \iint_R f(\vec r(u,v)) \left |\frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v}\right| dudv.
\end{equation*}
If we let \(f=1\text{,}\) then the surface area of \(S\) is simply
\begin{equation*}
\sigma = \iint_S d\sigma = \iint_R \left |\frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v}\right| dudv.
\end{equation*}
This definition tells us how to compute any surface integral. The steps are almost identical to the line integral steps.
Start by getting a parametrization \(\vec r\) of the surface \(S\) where the bounds form a region \(R\text{.}\)
Find a little bit of surface area by computing \(d\sigma =\left |\frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v}\right| dudv.\)
Multiply \(f\) by \(d\sigma\text{,}\) and replace each \(x\text{,}\) \(y\text{,}\) \(z\) with what they equals from the parametrization.
Integrate the previous function along \(R\text{,}\) your parameterization's bounds.
Exercise12.1.2
Consider the surface \(S\) given by \(z=9-x^2-y^2\text{,}\) for \(z\geq 0\text{.}\) A parametrization of this surface is
\begin{equation*}
\vec r(x,y) = (x,y,9-x^2-y^2), \text{ where } 9-x^2-y^2\geq 0.
\end{equation*}
(a)
Give a set of inequalities for \(x\) and \(y\) that describe the region \(R\) over which we need to integrate. The inequalities you give should be in a form that you can use them as the bounds of a double integral.
(b)
Find \(d\sigma = \left|\vec r_x\times \vec r_y\right|dxdy\text{.}\)
(c)
Set up the surface integral \(\iint_S d\sigma\) as an iterated double integral over \(R\text{.}\)
(d)
Convert the integral above to an integral in polar coordinates (don't forget the Jacobian).
Exercise12.1.3
Consider the surface \(S\) given by \(z=9-x^2-y^2\text{,}\) for \(z\geq 0\text{.}\) A different parametrization of this surface is
\begin{equation*}
\vec r(r,\theta) = (r\cos\theta,r\sin\theta,9-r^2), \text{ where } 9-r^2\geq 0.
\end{equation*}
(a)
Give a set of inequalities for \(r\) and \(\theta\) that describe the region \(R_{r\theta}\) over which we need to integrate. The inequalities you give should be in a form that you can use them as the bounds of a double integral.
(b)
Find \(d\sigma = \left|\vec r_r\times \vec r_\theta \right|drd\theta\text{.}\)
(c)
Set up the surface integral \(\iint_S d\sigma\) as an iterated double integral over \(R_{r\theta}\text{.}\)
Exercise12.1.4
Find, that is actually compute, the surface area of the surface \(S\) given by \(z=9-x^2-y^2\text{,}\) for \(z\geq 0\text{.}\) Do this by computing any of the integrals from the previous two problems.
Exercise12.1.5
(a)
If a surface \(S\) is parametrized by \(\vec r(x,y) = (x,y,f(x,y))\text{,}\) show that \(d\sigma = \sqrt{1+f_x^2+f_y^2}\ dxdy\) (compute a cross product).
(b)
If \(\vec r(x,z) = (x,f(x,z),z)\text{,}\) what does \(d\sigma\) equal (compute a cross product - you should see a pattern)?
(c)
Use the pattern you've discovered to quickly compute \(d\sigma\) for the surface \(x=4-y^2-z^2\text{.}\)
(d)
Set up an iterated double integral that would give the surface area of \(S\) for \(x\geq 0\text{.}\)
Exercise12.1.6
Consider the sphere \(x^2+y^2+z^2=a^2\text{.}\) We'll find \(d\sigma\) using two different parameterizations.
(a)
If you use the rectangular parametrization \(\vec r(x,y) = (x,y,\sqrt{a^2-x^2-y^2})\text{,}\) what is \(d\sigma\text{?}\) [Hint, use the previous problem.]
(b)
Why can this parametrization only be use if the surface has positive \(z\)-values?
(c)
You'll want to memorize this result.
If you use the spherical parametrization
\begin{equation*}
\vec r(\phi,\theta) = (a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi),
\end{equation*}
show that
\begin{equation*}
d\sigma = (a^2|\sin\phi|)d\phi d\theta= (a^2\sin\phi) d\phi d\theta,
\end{equation*}
where we can ignore the absolute values if we require \(0\leq \phi\leq \pi\text{.}\) Along the way, you'll show that
\begin{equation*}
\vec r_\phi\times \vec r_\theta = a^2\sin \phi (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi).
\end{equation*}
We can compute average value, centroids, center of mass, moments of inertia, and radii of gyration as before. We just replace \(dA\) with \(d\sigma\text{,}\) and all the formulas are the same.
Exercise12.1.7
Consider the hemisphere \(x^2+y^2+z^2=a^2\) for \(z\geq 0\text{.}\)
(a)
Set up a formula that would give \(\bar z\) for the centroid of the hemisphere. I suggest you use a spherical parametrization, as then the bounds are fairly simple, and we know \(d\sigma = (a^2\sin\phi) d\phi d\theta\) from the previous problem.
(b)
Compute the two integrals in your formula.
(c)
Set up an integral formula for \(R_z\text{,}\) the radius of gyration about the \(z\) axis, provided the density is constant.
Subsection12.1.1Flux across a surface
We now want to look at the flux of a vector field across a surface \(S\text{.}\) In the line integral section, we defined the outward flux of a vector field \(F\) across a curve \(C\) to be the line integral \(\ds \int_C \vec F\cdot \vec n ds\text{,}\) where \(\vec n\) is a normal vector point out of region enclosed by a curve \(C\text{.}\) When we want to find the flux of a vector field across a surface, we must state in which direction we want to compute the flux. We then must make sure that normal vector \(\vec n\) we choose to use actually points in the desired direction. The flux of a vector field \(\vec F\) across a surface \(S\) is the surface integral
\begin{equation*}
\text{ Flux } =\Phi
= \iint_S \vec F\cdot \vec n d\sigma
.
\end{equation*}
The next problem will help us simplify the computation of \(\vec nd\sigma\text{.}\)
Exercise12.1.8
For both parts of this exercise, the computations involved were actually done in previous problems. You just need to compile the information here.
Consider again the surface \(z=9-x^2-y^2\text{.}\)
(a)
(i)
Find a unit normal vector \(\vec n\) to the surface so that \(\vec n\) points upwards away from the \(z\)-axis. Make sure you explain how you know the normal vector you give is pointing upwards away from the \(z\) axis.
(ii)
State what \(d\sigma\) equals, as well as \(\vec n d\sigma\text{.}\)
(b)
(i)
Find a unit normal vector \(\vec n\) to the surface so that \(\vec n\) points downwards towards the \(z\)-axis. Make sure you explain how you know the normal vector you give is pointing downwards towards the \(z\) axis.
(ii)
State what \(d\sigma\) equals, as well as \(\vec n d\sigma\text{.}\)
In the problem above, we showed that \(\vec n d\sigma = \pm(\vec r_x\times\vec r_y)dxdy\) and that \(\vec n d\sigma = \pm(\vec r_r\times\vec r_\theta)drd\theta\text{.}\) We no longer need to find the magnitude of the cross product, but we must determine the correct sign to put on our cross product. This shows us that we can write flux as
\begin{equation*}
\text{ Flux } =\Phi
= \iint_S \vec F\cdot \vec n d\sigma
= \iint_{R_{uv}} \vec F\cdot (\pm \vec r_u\times \vec r_v) dudv
.
\end{equation*}
Exercise12.1.9
Complete either this challenge or the next exercise (same question, but with more help/walkthrough).
Consider the cone \(z^2=x^2+y^2\) and vector field \(\vec F = (2x+3y, x-2y, yz)\text{.}\) Set up an iterated integral that would give the flux of \(\vec F\) outwards (away from the \(z\)-axis) for the portion of the cone between \(z=1\) and \(z=3\text{.}\)
Exercise12.1.10
Consider the cone \(z^2=x^2+y^2\) and vector field \(\vec F = (2x+3y, x-2y, yz)\text{.}\) We want to set up an iterated integral that would give the flux of \(\vec F\) outwards (away from the \(z\)-axis) for the portion of the cone between \(z=1\) and \(z=3\text{.}\)
(a)
Start by parameterizing the cone by using a polar parametrization
\begin{equation*}
x=r\cos\theta, y=r\sin\theta, z=?.
\end{equation*}
(b)
Next obtain bounds for \(r\) and \(\theta\) that are constants.
(c)
Compute the normal vector and look at the third component to determine if it points up or down.
(d)
Finally just plug everything into the formula.
When the surface is flat, often you can determine the normal vector without having to perform any cross products. We'll now compute a flux of a vector field outwards across the 6 faces of a cube.
Exercise12.1.11