Question

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... :(

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

We basically go about this through substitution.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

Fine so far?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

yes

Answer 18

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

Answer 19

let me try using your notation,

Answer 20

is it

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

Yes, except you left out an 'a'.

Answer 25

that was very clear explanation right there.
bravo

Answer 26

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

Answer 27

Easy!

Answer 28

so the answer is

Answer 29

pi a^2

Answer 30

pia^2

Answer 31

Concur...

Answer 32

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

Answer 33

Great :)

Answer 34

oh, one question.

Answer 35

does volume work similarly ?

Answer 36

You would have to make suitable substitutions again.

Answer 37

gotcha