## yuki 5 years ago x = 2a*tan(t) and y = 2a*cos^2(t) what is the area in the first quadrant bounded by the above parametric equations, x=0 and x=2a? I know how to find an area under some curves in rectangular and polar coordinates, but not for parametric curves... :(

1. anonymous

Sorry, I'm half here, half not...distractions.

2. anonymous

So you don't know how these integrals are put together?

3. Yuki

It's been 7years since I took my last calculus class ;p

4. anonymous

It's similar to finding the area under the curve, $y=F(x)$for $a \le x \le b$

5. Yuki

take your time, I'm the one who's getting help.

6. anonymous

7. anonymous

We know that x = f(t), and so there will be alpha, beta such that a=f(alpha), b=f(beta). Also, $dx=f'(t)dt$

8. anonymous

So$A=\int\limits_{a}^{b}H(x)dx=\int\limits_{f(\alpha)}^{f(\beta)}H(f(t))f'(t)dt$

9. Yuki

by H(x) do you mean F(x) ?

10. anonymous

Since y=H(x), y=H(f(t))=g(t) here, so$A=\int\limits_{f(\alpha)}^{f(\beta)}g(t)f'(t)dt$

11. anonymous

Yeah, sorry...H=F here...sloppy accounting.

12. anonymous

We're assuming the curve is traced out at most once as t roams from its initial value to its last.

13. anonymous

Fine so far?

14. anonymous

Okay...I'll assume yes...and continue...

15. Yuki

are you saying,$\int\limits_{x(0)}^{x(2a)} y(t)*x'(t) dt$

16. anonymous

$g(t)=2a \cos ^2 t$and$f(t)=2a \tan t \rightarrow f'(t)=2a \sec^2 t$

17. anonymous

yes

18. anonymous

I'm used to my notation... :[

19. Yuki

let me try using your notation,

20. Yuki

is it

21. anonymous

$\therefore g(t)f'(t)=4a^2 \cos^2 t \sec^2 t=4a^2$

22. anonymous

The integral is now easy. You have to find the limits.

23. Yuki

$\int\limits_{0}^{\pi/4} 2acos^2(t)*2\sec^2(t) dt ?$

24. anonymous

Yes, except you left out an 'a'.

25. Yuki

that was very clear explanation right there. bravo

26. anonymous

So it's$4a^2\int\limits_{0}^{\pi/4}dt$

27. anonymous

Easy!

28. Yuki

so the answer is

29. anonymous

pi a^2

30. Yuki

pia^2

31. anonymous

Concur...

32. Yuki

awesome. I will try to solve it on my own again.

33. anonymous

Great :)

34. Yuki

oh, one question.

35. Yuki

does volume work similarly ?

36. anonymous

You would have to make suitable substitutions again.

37. Yuki

gotcha