Question

Calculus 3 help

Answer 1

What do you mean ? You don't see the attachment ?

Answer 2

There it is sorry

Answer 3

I was doing it on mobile

Answer 4

*

Answer 5

Here's a rendering of the region \(S\) (the red part of the cylinder between the planes \(z=0\) and \(z=x\) with \(0\le x\le3\)). (attached)

The way I read it, you're computing a surface integral, which means you're going to have to parameterize the surface \(S\). Since it's on a cylinder, using cylindrical coordinates looks like your best bet. Usually this is a conversion from \((x,y,z)\) to \((r,\theta,z)\), but just to not trip over identical symbols later on, I'll use \(\zeta\) (zeta) in place of the cylindrical \(z\), and \(t\) in place of \(\theta\) for brevity.

We can get the cylinder by setting \(x=3\cos t\) and \(y=3\sin t\), and \(z=\zeta\). We know that \(0\le z\le x\), so \(z\) must be non-negative. In cylindrical coordinates, this translates to \(0\le \zeta\le 3\cos t\). So to enforce \(\zeta\ge0\), we need to choose \(t\) such that \(\cos t\ge0\), and this occurs for \(|t|\le\dfrac{\pi}{2}\).

This is enough to parameterize the surface \(S\) with the set \(T\): \(T=\left\{(r,t,\zeta)~:~r=3,~|t|\le\dfrac{\pi}{2},~0\le \zeta\le3\cos t\right\}\). We will then be referring to each of \(x,y,z\) as functions of \(t\) and \(\zeta\), which we can represent with the vector function \(\mathbf x(t,\zeta)=\langle x(t,\zeta),y(t,\zeta),z(t,\zeta)\rangle\).

So, the integral is
\[\iint_S f(x,y,z)\,\mathrm dS=\iint_Tf(\mathbf x(t,\zeta))\left\|\frac{\partial\mathbf x(t,\zeta)}{\partial t}\times\frac{\partial\mathbf x(t,\zeta)}{\partial \zeta}\right\|\,\mathrm dt\,\mathrm d\zeta\]where
\[\begin{align*}
f(\mathbf x(t,\zeta))&=x(t,\zeta)^2+y(t,\zeta)^2+z(t,\zeta)^2\\[1ex]
&=9\cos^2t+9\sin^2t+\zeta^2\\[1ex]
&=9+\zeta^2
\end{align*}\]and
\[\begin{cases}
\dfrac{\partial\mathbf x(t,\zeta)}{\partial t}=\dfrac{\partial}{\partial t}\langle3\cos t,3\sin t,\zeta\rangle=\langle-3\sin t,3\cos t,0\rangle\\[2ex]
\dfrac{\partial\mathbf x(t,\zeta)}{\partial \zeta}=\dfrac{\partial}{\partial \zeta}\langle3\cos t,3\sin t,\zeta\rangle=\langle0,0,1\rangle
\end{cases}\\[2ex]
\implies\left\|\frac{\partial\mathbf x}{\partial t}\times\frac{\partial\mathbf x}{\partial\zeta}\right\|=\|\langle3\cos t,3\sin t,0\rangle\|=3\]So the integral reduces to
\[\iint_Tf(\mathbf x(t,\zeta))\left\|\frac{\partial\mathbf x(t,\zeta)}{\partial t}\times\frac{\partial\mathbf x(t,\zeta)}{\partial \zeta}\right\|\,\mathrm dt\,\mathrm d\zeta=3\int_{-\pi/2}^{\pi/2}\int_0^{3\cos t}(9+\zeta^2)\,\mathrm d\zeta\,\mathrm dt\]and the rest is easy to compute.

Answer 6

If you use the cylindrical surface area element \(dA = R ~ d z ~ d\theta \), and recognise that because \(z = x\) along that top plane, then i think you can grab \( 0 \le z \le 3 \cos \theta\) as your limit and go straight here:

\(\large \int\limits_{ \theta = - \pi/2}^{ \pi/2} \int\limits_{ z = 0}^{3 \cos \theta} (x^2 + y^2 + z^2) \cdot 3 ~ dz ~ d \theta \)

and because \(x^2 + y^2 = 9\) on the cylindrical surface, you reduce it down to:

\(\large 3 \int\limits_{ \theta = - \pi/2}^{ \pi/2} \int\limits_{ z = 0}^{3 \cos \theta} 9+ z^2 ~ dz ~ d \theta \)

which is the poor man's way of getting to holster's conclusion.

The thing that really confused me about this one is that the description of S looks to be a volume and not a surface, or put it another way, S is actually 3 surfaces enclosing a volume.

No doubt it's my bad, but it's because of the way the \(0 \le z \le x\) thingy at the end is written..... should it not actually read \(z = x\) so as to specify that we are on that top slanty surface alone and not looking at the total surface of an enclosed volume?!?!

Otherwise aren't we allowing z to float freely within that volume space?

PS it comes out as 198. I think :)

Answer 7

I was also confused at first about how to interpret it, but the double integration seemed to point to a surface integral. Hope we're right about that...

Answer 8

The area is the surface between two curves
\[
u(t)=\{3 \cos (t),3 \sin (t),3 \cos (t)\}\, -\frac \pi 2 \leq t \leq \frac \pi 2
\]
and
\[
v(t)=\{3 \cos (t),3 \sin (t),0\}\, -\frac \pi 2 \leq t \leq \frac \pi 2
\]
I will upload some graphs to show that

Answer 9

Here are the two curves

Answer 10

Here is the surface joining them

Answer 11

To be able to draw this curve, I had to parametrize this area,

This quite easy\[
q(s,t)=u(t)+ s(v(t)-u(t)) =\{3 \cos (t),3 \sin (t),3 \cos (t)-3 s \cos (t)\}
\]

Answer 12

\[
q(s,t)=u(t)+ s(v(t)-u(t)) =\\\{3 \cos (t),3 \sin (t),3 \cos (t)-3 s \cos (t)\}
\]

Answer 13

s between 0 and 1

Answer 14

Set\[
\{x(\text{s$\_$},\text{t$\_$}),y(\text{s$\_$},\text{t$\_$}),z(\text{s
$\_$},\text{t$\_$})\}=\\\{3 \cos (t),3 \sin (t),3 \cos (t)-3 s \cos
(t)\}
\]

Answer 15

Compute
\[
\text{AreaConversionFactor}=\text{Norm}\left[\left[\frac{\partial
q(s,t)}{\partial s}\times \frac{\partial q(s,t)}{\partial
t}\right]\right]=9 \cos (t)
\]

Answer 16

Finally
\[
\int _0^1\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}9 \cos (t)
f(x(s,t),y(s,t),z(s,t))dtds=198
\]

Answer 17

This does agree with the previous method, which may be easier than mine
\[
3 \int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \left(\int_0^{3 \cos
(\theta )} \left(z^2+9\right) \, dz\right) \, d\theta =198
\]

Answer 18

The good thing about my method, is the possibility to plot the surface directly