A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 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... :(
anonymous
 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... :(

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, I'm half here, half not...distractions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you don't know how these integrals are put together?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's been 7years since I took my last calculus class ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's similar to finding the area under the curve, \[y=F(x)\]for \[a \le x \le b\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take your time, I'm the one who's getting help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We basically go about this through substitution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We know that x = f(t), and so there will be alpha, beta such that a=f(alpha), b=f(beta). Also, \[dx=f'(t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So\[A=\int\limits_{a}^{b}H(x)dx=\int\limits_{f(\alpha)}^{f(\beta)}H(f(t))f'(t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0by H(x) do you mean F(x) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since y=H(x), y=H(f(t))=g(t) here, so\[A=\int\limits_{f(\alpha)}^{f(\beta)}g(t)f'(t)dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, sorry...H=F here...sloppy accounting.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We're assuming the curve is traced out at most once as t roams from its initial value to its last.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...I'll assume yes...and continue...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you saying,\[\int\limits_{x(0)}^{x(2a)} y(t)*x'(t) dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[g(t)=2a \cos ^2 t\]and\[f(t)=2a \tan t \rightarrow f'(t)=2a \sec^2 t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm used to my notation... :[

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me try using your notation,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\therefore g(t)f'(t)=4a^2 \cos^2 t \sec^2 t=4a^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The integral is now easy. You have to find the limits.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/4} 2acos^2(t)*2\sec^2(t) dt ?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, except you left out an 'a'.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that was very clear explanation right there. bravo

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So it's\[4a^2\int\limits_{0}^{\pi/4}dt\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome. I will try to solve it on my own again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does volume work similarly ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You would have to make suitable substitutions again.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.