mixedmath

2011-04-14 davidlowryduda

Recall that $ \vec{F} (x,y,z) = P \hat{i} + Q \hat{j} + R \hat{k}$ on a curve $ C$ parameterized by $ \vec{r} (t) = x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k}; a \leq t \leq b$ has a nice relationship with our line integrals. That is:

$$ \int_C \vec{F} \cdot d \vec{r} = \int^b_a (P \hat{i} + Q \hat{j} + R \hat{k}) \cdot (x' \hat{i} + y' \hat{j} + z' \hat{k})$$ $$ = \displaystyle\int^b_a Px'dt + \displaystyle\int^b_a Qy'dt + \displaystyle\int^b_a Rz'dt$$ $$ = \displaystyle\int_C Pdx + Qxy + Rxz$$

This is so often forgotten!

2011-04-14 davidlowryduda

Whoops! That last line is:

$ \displaystyle\int_C Pdx + Qdy + Rdz$

2011-04-15 Anony

Is this only true for gradients?

2011-04-15 davidlowryduda

It's true in general. At the least, it's true for this course as we only consider well-behaved functions and relatively smooth curves.

2011-04-17 Dhruv

Hey David...Could you please give an explained solution to question no.45 on pg 888 in the book. Thanks

2011-04-17 Dhruv

I did not know you were the SGA rep. MATH dept. until I read your name in the following article:

http://www.nique.net/news/2011/04/15/uhr-gss-send-budget-to-conference-committee/

2011-04-17 davidlowryduda

Yepyep! And I've just spent the last 4 hours or so on the Conference Committee for the Budget. It's about $4.3 million or so. Exciting? Sort of. Exhausting? certainly.

2011-04-17 davidlowryduda

Sure. So the problem is to find the volume of the solid in the first octant bounded by the cylinders $ x^2 + y^2 = a^2$ and $ x^2 + z^2 = a^2$.

I will attempt to post a picture here: http://www.math.tamu.edu/~tkiffe/calc3/newcylinder/cylinderb.jpeg

The easiest way I see to do this is with a standard double integral in rectangular coordinates. We restrict ourselves to only positive x,y,z values. So I will integrate the function $ f(x,y) = \sqrt{a^2 - x^2}$, gotten from the horizontal cylinder $ x^2 + z^2 = a^2$.

That takes care of the height. Over what 2 dimensional domain? Well, we need only consider positive x and y values, and we are limited only be the vertical cylinder. So if I choose to integrate with respect to y first, we get the following:

$ Volume = \displaystyle\int^a_0 \displaystyle\int^{\sqrt{a^2 - x^2}}_0 \sqrt{a^2 - x^2} dydx$

And this integrates nicely and easily to $ \dfrac{2a^3}{3}$.

How's that?

2011-04-18 Patrick

Hey David, can you perhaps post a graph of number 37 from 17.9 (page 930)? I'm having a hard time understanding why the integration is split into two parts.

Thanks!

2011-04-18 davidlowryduda

My first embedded image! The idea is that we use spherical coordinates, and so we care about the bounding radii. Here, we can clearly see that there are parts where the boundary is on one sphere as opposed to the other.

How's that?

As a reminder, you should do all the practice he recommends, not just the 4 problems he puts up for the test. I should also note that once I have approved a comment, your comments should appear freely.

(Later edit: the comment system for this site has completely changed)

2011-04-18 davidlowryduda

My first embedded image and I used the wrong one!

Attempt number 2!

2011-04-18 Dhruv

Hey David...what r ur office hours, and where is ur office?

2011-04-19 davidlowryduda

I have one office hour tomorrow. It's at 11 am and I hold it in the TA lounge, something like Skiles 233 (or maybe 230, 236, something right around there). If you have trouble finding it, ask someone in the hall and they can direct you to it.