Question

Can someone explain it? work through it with me.
http://imgur.com/7zoYyBB

Answer 1

Wow. Have you done any of it so far?

Answer 2

Start with (a): you will be trying to get rid of the variable t, by doing operations on
y=3*cos(t) and x=9*sin(2*t) You will need the help of some trigonometric identities. Know anyplace to find them? we need one that will express sin(2*t) with a function of t (I think)

Answer 3

@mjdennis I don't think it requires changing the Parametric equation to Cartesian equation. If that's what u r trying to say.

Answer 4

because the parametric equation can be integrated easily as well.

Answer 5

what I don't understand is that the area under the graph above x-axis is positive and below x-axis is negative when found using integrals. but the graph is symmetrical below and above x-axis. Wouldn't the definite integral in that period be zero? instead of the shaded region.

Answer 6

@HollowDensity it does; read the question. Part a.

Answer 7

yea. sorry for part a. it does. but i'm having trouble with Part b

Answer 8

You can probably use your answer from part a first, to find the area...

Answer 9

\[y^2 = 4x^2(9-x^2)\]

Answer 10

That makes b) really easy.

Answer 11

that is the cartesian, i found.

Answer 12

Yes, in (b) we will go back to parametric. In (a) it pretty clearly says "Find the Cartesian equation..." if you do not want to solve that part, say so.

Answer 13

Use the rectangular form for part b. It's a simple substitution integral.

Answer 14

@mjdennis Yea Sorry, my bad about that. i was looking particularly for part b

Answer 15

@agent0smith ??. i dont understand. did u look at the graph?
"what I don't understand is that the area under the graph above x-axis is positive and below x-axis is negative when found using integrals. but the graph is symmetrical below and above x-axis. Wouldn't the definite integral in that period be zero? instead of the shaded region."

Answer 16

I am too slow answering today. @agent0smith better here.

Answer 17

\[y = \sqrt{4x^2(9-x^2)}\]
you only need the area in the first quadrant. Use the positive square root

\[\large \int\limits_{0}^{?}2x \sqrt{9-x^2}\]

Answer 18

yea that makes some sense now. wait let me try it. But the required answer should be in specified form.

Answer 19

Once you have that value (the upper limit looks like pi/2 btw, and i guess i forgot dx lol) then you can equate it to \[A \int\limits_{0}^{\pi/2}\sin 2t \sin t dt\]
and use an identity for sin2t to make this into an easy integral, too.

Answer 20

Oh yeah you can demonstrate b) pretty easily just by using integrals of parametric equations http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea.aspx

Answer 21

i saw it in solution manual and this is how they did it.
http://imgur.com/xru3Aui

Answer 22

Yeah, see the link above. It's pretty easy really.

Answer 23

but it again confuses me. because that would give sum of area above and below x-axis

Answer 24

The definite integral because of how its defined

Answer 25

No, 0 to pi would give the area above and below, wouldn't it?

All you need to look at on that link is:
http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea_files/eq0001MP.gif
and
http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea_files/eq0013MP.gif

Answer 26

okay let me take a look

Answer 27

i took a look. i can do this. its integration of parametric equation.

Answer 28

Just looking at the graph, and knowing the graph shows from 0 to 2pi, should tell you that each "quarter" of the graph is equal to a time period of pi/2.

Answer 29

@agent0smith bulls-eye, that last line just dissolved the confusion. thanks a lot

Answer 30

Heh, you're welcome. I didn't bother looking at the equations and trying to do math to figure out when the curve first touches the x-axis. Sometimes being lazy saves time.

Answer 31

ikr