Author Topic: TT's Maths Thread (Read 119594 times) Tweet Share

TrueTears · « **Reply #1095 on:** November 03, 2010, 04:13:13 pm »

1. A sphere with radius 2 ( $x^2+y^2+z^2=4$ ) has its cylindrical core of radius 1 removed ( $x^2+y^2=1$ ), what is the volume of the resultant solid?

a) Use cylindrical coordinates to find the volume of the resultant solid.
b) Use spherical coordinates to find the volume of the resultant solid.

For a) I just used cylindrical coordinates to find the volume of the cylinder, then used $V = \frac{4}{3}\pi r^2$ which is the volume of the sphere then minus the volume of the cylinder from it.

For b) I used spherical coordinates to find the volume of the sphere, then used $V = \pi r^2 h$ which is the volume of the cylinder, then used the volume of sphere minus the volume of cylinder.

Now what I am wondering is... is that the right way to interpret the question? Or does the question mean use cylindrical coordinates ONLY to work out the volume? If so... how do I do that? Cause I thought as long as I used cylindrical coordinates to find the volume that's fine.

Same goes for b)

2. $\mathbf{F} = (x+y)i+(y+z)j+(z+x)k$ . Verify the Divergence Theorem by showing $\int \int \int_E \Delta . \mathbf{F} dV = \int \int_S \mathbf{F}.\mathbf{n} dS$ where S is the union of the surfaces $z=0$ and $z = 4-x^2-y^2$ and E is the solid encompassed by those 2 surfaces.

Now I can evaluate the LHS which in spherical coordinates is given by: $\int_0^{\frac{\pi}{2}} \int_0^{2\pi} \int_0^4 3p^2\sin(\phi)dp d\theta d\phi = 128\pi$ (btw is this triple integral right? As in did I set it up right, not the answer

)

Now how do I evaluate the RHS for the surface integral over the surface $z = 4-x^2-y^2$ (without a calculator)... it's almost impossible to do by hand after you compute $\mathbf{F}(\mathbf{r}(u,v)).(\mathbf{r}_u \times \mathbf{r}_v)$ where u and v are x and y respectively.

3. Evaluate $\int_1^2 \int_{\frac{1}{y}}^{\sqrt{y}} e^{\sqrt{xy}}dxdy$ by making the transformation $x = \frac{u}{v}$ and $y = uv$

I swear you can NOT do this question without a CAS.

After sketching on the u-v plane (with u as the vertical axis) we get the region to be the region between $u = v^3, u = \frac{2}{v}$ and $u = 1$ (I've checked this many times)

Now the Jacobian is $\frac{2u}{v}$ and after making the transformations we can evaluate the integral in the u-v plane with: $\int_1^{2^{\frac{3}{4}}} \int_{u^{\frac{1}{3}}}^{\frac{2}{u}} 2ue^u dvdu$

Now it's impossible to integrate that without a calculator.

Then we can try reverse the order of the integral so we get a dudv instead of dvdu, it's also unable to be integrated by hand (try it yourself). So is there ANYWAY to do this question by hand?

/0 · « **Reply #1096 on:** November 03, 2010, 06:30:18 pm »

For 1, how did you use $V=\pi r^2h$ for the cylinder, since it doesn't have constant height? Did you use it with another integral? If you were to use spherical coords to find the volume of the cylinder it would go like:

$V=\underbrace{\pi (1)^2(2\sqrt{3})}_{\mbox{volume of `cylindrical prism' bit}}+2\underbrace{\int_0^1\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{6}} r^2\sin\theta\,d\theta\,d\phi\,dr}_{\mbox{volume of caps}}$ , where $\phi$ is the angle with the x-axis and $\theta$ the angle with the z-axis.

With 2, I'm not sure if you can use spherical coordinates, since it's a parabola... if you use cylindrical then you have

$\int_{0}^2\int_0^{4-s^2}\int_{0}^{2\pi} 3s\,d\phi\,dz\,ds$ ( $s=x^2+y^2$ )

Or in cartesian,

$\int_{-2}^2\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}\int_0^{4-x^2-y^2} 3\,dz\,dx\,dy$

The surface is:

$\mathbf{r}=x\mathbf{i}+y\mathbf{j}+(4-x^2-y^2)\mathbf{k}$

$\mathbf{r}_x=\mathbf{i}-2x\mathbf{k}$

$\mathbf{r}_y=\mathbf{j}-2y\mathbf{k}$

$\mathbf{r}_x\times \mathbf{r}_y=2x\mathbf{i}+2y\mathbf{j}+\mathbf{k}$

$\mathbf{F}\cdot(\mathbf{r}_x\times\mathbf{r}_y)=2x(x+y)+2y(y+z)+z+x$

So the integral is

$\int_{-2}^2\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} 2x(x+y)+2y(y+4-x^2-y^2)+4-x^2-y^2+x\,dx\,dy$

This is horrible... lol

But switching to polar coordinates may help again:

$\int_0^2\int_0^{2\pi} (2s\cos{\phi}(s\cos\phi+s\sin\phi)+2s\sin{\phi}(s\sin\phi+4-s^2)+4-s^2+s\cos\phi)s\,d\phi\,ds=\int_0^2\int_0^{2\pi} (s^2+2s^2\cos\phi\sin\phi+8s\sin\phi-2s^3\sin\phi+4+s\cos\phi)s\,d\phi\,ds$

Now $\int_0^{2\pi}\sin\phi\,d\phi=\int_0^{2\pi}\cos\phi\,d\phi=\int_0^{2\pi}\cos{\phi}\sin\phi\,d\phi=0$ , so you're left with the slightly more manageable

$\int_0^2\int_0^{2\pi} s^3+4s\,d\phi\,ds=24\pi$

Still it's absurd they'd ask you to do that in a non-calc exam... I had to go back and fix my working on this question so many times, and I had a calculator -.- lol

I'm still stuck with the 3rd one though... like u said i can't find an antiderivative for it

TrueTears · « **Reply #1097 on:** November 03, 2010, 06:48:28 pm »

Thanks heaps dude, hmm yeah I thought you couldn't use spherical since it was a parabola... I had the cylindrical coordinates too lol I wrote both...

then I spent like 20 mins on the RHS which like you said is insane without a calculator ><

And yeah I'm pretty sure the third one is a mistake in the exam, if you changed the terminal from rt[y] to y then it works out nicely with the rt[y] i don't think you can find an antiderivative.

also there were so many mistakes in the exam, had to tell the invigilators to fix em sigh... so pissed

The cylinder does have a constant height doesn't it?

I interpreted the cylindrical core as in a cylinder as the "core" of the sphere, which intersects the sphere on the planes $z = -\sqrt{3}$ and $\sqrt{3}$

so the height is just $2\sqrt{3}$

so eg, for b) $\int \int \int_E 1 dV$ in spherical coordinates, where E is the solid of a sphere with centre (0,0,0) and radius 2, then minus the $V=\pi r^2h$

/0 · « **Reply #1098 on:** November 03, 2010, 06:59:20 pm »

Hmmm I guess there are different interpretations of the questions... sounds like the exam wasn't well written :/

My waves and optics exam on monday will probably be similar, sigh... -.-

TrueTears · « **Reply #1099 on:** November 03, 2010, 07:00:00 pm »

Quote from: /0 on November 03, 2010, 06:59:20 pm

Hmmm I guess there are different interpretations of the questions... sounds like the exam wasn't well written :/

My waves and optics exam on monday will probably be similar, sigh... -.-

how did you interpret it? can u kinda sketch it? LOL but do u get my interpretation?

/0 · « **Reply #1100 on:** November 03, 2010, 07:05:02 pm »

Hmmm I interpreted it like this:

But I get your interpretation as well, since it isn't really clear what they mean by 'core'

TrueTears · « **Reply #1101 on:** November 03, 2010, 07:35:46 pm »

Ahh yeah I thought about it like that too in the exam, but I still stuck with my current interpretation, sigh, they really need a picture to show what they really want, or else there are too many interpretations... the reason why I rejected how you interpreted in the exam was because see the blue shaded part? the "cap" doesnt make the core a cylinder anymore technically right?

/0 · « **Reply #1102 on:** November 03, 2010, 08:48:17 pm »

Quote from: TrueTears on November 03, 2010, 07:35:46 pm

the "cap" doesnt make the core a cylinder anymore technically right?

good point

TrueTears · « **Reply #1103 on:** November 03, 2010, 09:17:48 pm »

yeah stupid exam, sigh... lol oh wells

gl for your exams man, how many have you done and how many left??

Thanks for all your help again man, really appreciate it :smitten:

/0 · « **Reply #1104 on:** November 03, 2010, 09:52:21 pm »

lol gl for the rest of your exams too
for me the pain is yet to come, I've got exams on monday 8th, then 16th, 17th, and 18th...

I've been really lazy so it's gonna be cram central

TrueTears · « **Reply #1105 on:** November 03, 2010, 11:57:14 pm »

haha thanks

oh I see, I finish next week on the 9th can't wait

Only got commerce exams left, all my maths exams are done! (I loved my pure maths exam btw, my first ever pure maths exam <3)

then it's happy-fun-maths-time as soon as exams end!!

are you coming back to Melbourne for holidays after exams end? we gotta catch up

jimmy999 · « **Reply #1106 on:** November 10, 2010, 10:43:27 pm »

Quote from: TrueTears on November 03, 2010, 04:13:13 pm

$\int_1^{2^{\frac{3}{4}}} \int_{u^{\frac{1}{3}}}^{\frac{2}{u}} 2ue^u dvdu$

Now it's impossible to integrate that without a calculator.

Then we can try reverse the order of the integral so we get a dudv instead of dvdu, it's also unable to be integrated by hand (try it yourself). So is there ANYWAY to do this question by hand?

Chuck this integral into Wolfram Alpha without the limits of the u variable. You'll find it that it gives you the gamma function instead of an exact antiderivative. When doing this question, I got as far as I could, then said that part of the integral couldn't be solved using elementary functions. In all fairness, they have to disregard the final bit of that question when marking the exam as it's not doable without a calculator.

TrueTears · « **Reply #1107 on:** November 11, 2010, 01:44:45 pm »

Yeah, I'm really pissed at the exam because I spent like 10-15 mins on that question thinking wdf, it said "EVALUATE the integral" meaning you should be able to get an exact answer and you can't do it without a CAS. Wasted so much time.

TrueTears · « **Reply #1108 on:** November 15, 2010, 01:06:19 pm »

Show that if A and B are invertible matrices of the same size, then AB is invertible and $(AB)^{-1} = B^{-1}A^{-1}$

How do I prove this?

I was thinking that if we can show $(B^{-1}A^{-1})(AB) = (AB)(B^{-1}A^{-1}) = I$ then we are done... but how?

m@tty · « **Reply #1109 on:** November 15, 2010, 01:12:21 pm »

Quote from: TrueTears on November 15, 2010, 01:06:19 pm

Show that if A and B are invertible matrices of the same size, then AB is invertible and $(AB)^{-1} = B^{-1}A^{-1}$

How do I prove this?

I was thinking that if we can show $(B^{-1}A^{-1})(AB) = (AB)(B^{-1}A^{-1}) = I$ then we are done... but how?

$(AB)^{-1}(AB)=B^{-1}A^{-1}AB=B^{-1}IB=B^{-1}B=I$

It's invertible because it is square.

And as seen above $(AB)^{-1}$ must equal $B^{-1}A^{-1}$ because otherwise you won't get down to I.

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