c4 parametric question

Hello, I was just feeling a bit stuck on this question ...I would appreciate it if u could help me on it ... thanks!

There is this parametrics question : http://www.examsolutions.net/maths-r...tutorial-1.php

I did the following and just wanted to check with u if I have done it right:

x=2cost y=sin2t

dx/dt= -2sint

so integral sign (sin2t) (-2sint)

then [sin(3t) / (3) - sint ] and for the limits I got t= 90 degrees and 0 . I subbed them in but got -4/3....what do i do next? am i going the right way is this one of those questions where the limits will swap around because we surely cannot have a negative area!?
thanks laura

you can have negative area, it just shows the area is below the x axis

so integral sign (sin2t) (-2sint)

then [sin(3t) / (3) - sint ]

Link is not working

what did you do there

Are you looking to integrate $-2\displaystyle \int \sin 2t \sin t dt$

yes and then i got

sin (3t) / 3 - sint

Then I subbed in the values of t = 90 degrees and t=0 and ended up with -4/3 .

On the web link, they have highlighted the top portion (it is an infinity shaped graph and they want us to work out the area of the right hand side blob (top of it above the x axis)) . I thought that when we use this method, we would find out the area of the whole of the right hand side blob (not just the top bit)

Something like this I'd imagine.

Your integral should have the "lower x value" at the bottom, and "larger x value" at the top, so should be from t=pi/2 at the bottom to t=0 at the top.

I get $-\dfrac{4}{3}[\sin^3t]$

But, yes I get $-\dfrac{4}{3}$ too assuming that is 90 at the top and 0 at the bottom

do you have a link to the question that works

wow , it is impressive how u produced this diagram! i know u would probably feel a bit angry if I ask this question again :
but
1) is this a question where the limits swap around?
2) so am I right or wrong?
3) when we do this method and sub in 90 and 0, do we get area of that whole right hand side blob?
4) is it ok to use 90 and 0 instead of radians version? thanks

If ghostwalker is correct then there is no need to swap anything

The limits would be 90 to 0 rather than 0 to 90

So the answer is $\dfrac{4}{3}$

wow , it is impressive how u produced this diagram! i know u would probably feel a bit angry if I ask this question again :
but
1) is this a question where the limits swap around?
2) so am I right or wrong?
3) when we do this method and sub in 90 and 0, do we get area of that whole right hand side blob?
4) is it ok to use 90 and 0 instead of radians version? thanks

so is my integration right sin3t / 3 -sint

ok if 90 is at the top of the integral sign thing surely we get -4/3 isnt it so"