Hope you do not mind explaining. In regards to "http://www.youtube.com/watch?v=nW2GD3pYsXY&feature=related" on triple integral and volume, at 06.48min, may I know how did he let x = 3sin ().

is there a formula to it?

 

It's the definition of the sine of an angle.

In the depicted triangle we have:
\(\sin(\theta)=\frac x3\)

Is that clear?

 

he drew the triangle out because he needed to solve the integral of "8*sqrt(9-x^2)"....

so how did he derive and get that triangle just from looking at this equation? or maybe can you explain and solve the integral of "8*sqrt(9-x^2)"?

also, when can i apply this formula? and how would i know that "sqrt(9-x^2)" is on the adjacent of the triangle? why can't it be on the opposite?

 

he drew the triangle out because he needed to solve the integral of "8*sqrt(9-x^2)"....

so how did he derive and get that triangle just from looking at this equation? or maybe can you explain and solve the integral of "8*sqrt(9-x^2)"?

also, when can i apply this formula? and how would i know that "sqrt(9-x^2)" is on the adjacent of the triangle? why can't it be on the opposite?

He knew to do this from experience. This is a well known trig substitution method. This would not be obvious to a student new to the subject.

The way to see the logic is to notice that sqrt(9-x^2) looks a little like the Pythagorean theorem a^2+b^2=c^2 which leads to b=sqrt(c^2-a^2). So, it's natural to think of a hypotenuse of length c=3.

Then, one should know the definition of sine and cosine etc. The sine is the ratio of the length of the opposite side of the angle to the length of the hypotenuse. The cosine is the ratio of the length of the adjacent side of the angle to the length of the hypotenuse. The tangent is the ratio of the length of the opposite side of the angle to the length of the adjacent side.

Note that he definitely takes a very long route to solve the problem. Many of the integrals he creates along the way are obviously zero by symmetry. Also the entire problem can be solved in your head without even needing integrals. But, clearly he is trying to teach a methodical method. All of the techniques he shows are things you should strive to learn and know off the top of your head. The best way to learn all this is to keep doing problems.

 

He realized that the term
\(\sqrt{9-x^2}\)
would be problematic.

So he decided to substitute it with something that would cancel out the root.
By letting \(x=3\cdot \sin (\theta)\)
you have:
\(\sqrt{9-x^2}\rightarrow \sqrt{9-9\cdot \sin^2 (\theta)}
=\sqrt{9 \cdot (1-sin^2 (\theta))}
=3\cdot \sqrt{cos^2(\theta)}
=3 \cdot |cos (\theta)|
=3 \cdot cos(\theta)
\)

The last step is possible because
\(
-3\leq x \leq 3 \leftrightarrow
-3 \leq 3 \sin (\theta) \leq 3 \leftrightarrow
-1 \leq \sin (\theta) \leq 1
\)
and for that area there is a θ that makes also cos(θ) positive.
That may lead to a loss of an imaginary solution, but since this a volume (hence real) problem, this is ok.

The triangle was only used to visualize the θ.

Those tricks aren't commonly found in Mathematical Analysis books, and cannot be formulated into rules. It's just experience from a point forward.

 

are there any alternative to solve this integration?

 

are there any alternative to solve this integration?

Generally, you have the following options.

1. Creative thought sometimes leads to a brilliant solution
2. Integral Tables allow you to solve without thinking
3. Computer based symbolic processors solve for you (e.g. Mathematica, Maple, Maxima, muPad etc.)


In this case, you could probably solve by power series using the binomial series expansion, but that is cumbersome.

Also, don't forget that (unlike this example) some ingtegrals are not solvable by any techniques, and must be calculated numerically..

 

i see. thank you very much.