In the past, I have talked about how good a supplemental source of information the Khan Academy is. Again, it is supplementary. But it seems to have lots of fully worked and fully explained examples of the concepts of chapters 17 and chapter 18 (sections 1 through 4) – the topics for your next exam. I have placed the relevant links below.

Double Integrals

Triple Integrals

Line Integrals

Clever Line Integrals

As always, if you have any questions let me know. I will be hosting a review session in the Math Lab at 5 - come prepared and with questions. I suspect we'll be focusing on the iterated integrals of Chapter 17. Good luck!

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Comments (13)2011-04-14 davidlowrydudaRecall 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 davidlowrydudaWhoops! That last line is:

$ \displaystyle\int_C Pdx + Qdy + Rdz$

2011-04-15 AnonyIs this only true for gradients?

2011-04-15 davidlowrydudaIt'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 DhruvHey David...Could you please give an explained solution to question no.45 on pg 888 in the book. Thanks

2011-04-17 DhruvI 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 davidlowrydudaYepyep! 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 davidlowrydudaSure. 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 PatrickHey 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 davidlowrydudaMy 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 davidlowrydudaMy first embedded image and I used the wrong one!

Attempt number 2!

2011-04-18 DhruvHey David...what r ur office hours, and where is ur office?

2011-04-19 davidlowrydudaI 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.