Simple Integration

[Chewy0087]

1. The problem statement, all variables and given/known data

http://i27.photobucket.com/albums/c171/Chewbacc0r/question.jpg


3. The attempt at a solution

there are a number of things i've had trouble with here,

the first part i got fine, bieng;

y^2 = 4x^2 ( 9 - x^2) , which is correct according to the answers, next

the 'show that' question, now i had no problem getting it into the right form using the;

\int f(t)\frac{dx}{dt}dt rule, so i put it in like so;

f(t) = 9 sin (2t) and \frac{dx}{dt} = -3 sin t,

so i put them together to get, \int -27 sin(2t) sin(t) , so A, the constant, would be minus 27...wrong! it simply says 27 in the back, so i thought, hey well maybe it's just an error, but i have a sneaking suspicion i might be wrong aswell so i would appreciate clarification on that one aswell.

next, "find the value of this integral".

i don't know but this is a hard, hard integral, and we havn't even covered integration by parts or any formal integration aside from the \frac{n^(n+1)}{n+1} stuff, I have however done "some" differential equations in my own time and am somewhat familiar with integration, i can wade my way through it sometimes however this one strikes me as very hard, i thought, well maybe i can break down the sin (2t), giving me,

\int cos(t) - (cos(t))^3 however from here I am also stumped....

also, it says to integrate between \frac{\pi}{2} and 0, why \frac{\pi}{2} i wonder? intuitiveley i looked at the equation and expanded the 9 - x^2 showing that the roots were x = 3 & x = -3, so if you put 3 (clearly) into the x equation to work out the value for t, 3 = 3 cos (t) -> 1 = cos (t) t = 0, 2[tex]/pi[tex] , so why \frac{\pi}{2}?

i would be much obliged if someone could give some advice on this one!

however i did look up the answer for this so as to complete the next part of the question, which was 18, and was able to do the last question

thanks guys i realise it's a long question but i think it's just me and it might only take someone good at maths 2 minutes :P, sorry again for the huge post and thank you

[rock.freak667]

Well it looks to me that the -ve sign would say that you re finding the area of the opposite side that is shaded (the part under the x-axis)

For the integral try using sin2t=2sintcost and then using a u substitution

[CompuChip]

For the second one, note that it is actually

\int f(t) \left|\frac{dx}{dt}\right| dt.

Geometrically it also makes sense that A > 0: the integral has a geometrical meaning as surface, which is always positive.

For the third question, you might find it helpful that
\sin(2t) = 2 \sin(t) \cos(t)
and use a substitution of variables.

(If the latter doesn't mean anything to you: try writing the integrand as a derivative f'(t) of some function f(t)).

[Fightfish]

By compound angle formula, (sin 2t)(sin t) = 1/2 ( cos (t) - cos (3t) ). That's an easily integrable form

[Chewy0087]

hmm thanks alot for the clearing up on the 2nd and 3rd question however i'm still wondering about the integrating between

pi/2 and 0 has come from, as on the x-axis you're integrating between 3 & 0, meaning 2pi and 0 for t....

[CompuChip]

For t = 0, the curve passes through (x, y) = (3, 0) and runs counter-clockwise. The upper integration limit is the time instant where it passes through the origin. When you solve for x = 0, y = 0, you will find that the first solution is t = pi / 2 (remember, you only need to integrate over the time interval in which the curve encloses the shaded area).

[Fightfish]

hmm thanks alot for the clearing up on the 2nd and 3rd question however i'm still wondering about the integrating between pi/2 and 0 has come from, as on the x-axis you're integrating between 3 & 0, meaning 2pi and 0 for t....
x = 3 cos t adopts the range of 0 to 3 for t having a range of 0 to pi/2. While other ranges appear to satisfy this condition, we must ensure that y= 9 sin 2t must be positive in that range of t chosen as well.

[Chewy0087]

Ahh i see, so the 'origin' for t is actually (3,0) ? :OO

is that a general rule for all parametric forms - finding out the (x,y) when it's 0 and integrating FROM that point?

much obliged, this assistance is top quality

[CompuChip]

No. In principle, what you want to do is calculate the area under a part of the curve y^2 = f(x). You need to find out which part that is. When you plug in t = 0 in the equation, you get (x, y) = (3, 0), which is the point all the way on the right on the x-axis. When you look at the shaded area, you see that you have to integrate up to the point where the graph crosses the x-axis (which, in this case, happens to be the origin). So you need to find which t belongs to this point. I suggest you do the calculation yourself and check that it gives t = pi / 2.

[Chewy0087]

No. In principle, what you want to do is calculate the area under a part of the curve y^2 = f(x). You need to find out which part that is. When you plug in t = 0 in the equation, you get (x, y) = (3, 0), which is the point all the way on the right on the x-axis. When you look at the shaded area, you see that you have to integrate up to the point where the graph crosses the x-axis (which, in this case, happens to be the origin). So you need to find which t belongs to this point. I suggest you do the calculation yourself and check that it gives t = pi / 2.

that's great, much obliged sir