Question

Find the area of the region enclosed by x(t)=t^2-2t, y(t)=sqrt(t), and the y-axis.

Answer 1

now i got that to be the intergral from 1 to 0 of sqrt(t) * (2t-2), not sure if thats correct

Answer 2

What you want to do is take the two points of intersection, in this case 0 (because it says the y-axis) and 1/4 (1+ sqrt(5))^2
Those will be your limits.
The integral isn't very straight forward either, since they arent in the same quadrant the whole time.
from 0-2 x(t) is below the x axis so you need to add that amount to integral y(t)
but after 2 you need to subtract the area under it from integral y(t)

Answer 3

hmmmmmm not sure i agree with you. this is a parametric calculus equation. you do not look for points of intersection for limits of integration I was taught. The limist come from limits of t which are where t = 0 and t=1. I am not sure, what are your qualifications, are you a math teacher?

Answer 4

So, you need to do something like:
\[\int\limits_{0}^{2}(\sqrt t+(t^2-2t) dt +\int\limits_{2}^{1/4 (1+\sqrt 5)^2} (\sqrt t-(t^2-2t))dt\]

Answer 5

Finding the point of intersection for the limits allows you to know the area that is enclosed.
Otherwise you are just enclosing it from 0 - 1 and not from the actual region that they create.
I'm not a math teacher, but when I was in grad school I taught a lot of math/physics/computational biology

Answer 6

As for the addition of the integrands (and integrals) it makes logical sense, you are finding the area under sqrt x adding it to the area under x-axis for the poly and then adding that to the area under sqrt x and subtracting that from the area under the poly (since it isnt enclosed if it is under both)

Answer 7

But actually, you need to take the abs of the poly in the first integral, since it will be a negative value... or I suppose that means you dont need to split it and you can take the whole thing as one integral and subtract the parts

Answer 8

im stilll super confused about the limits of integration?

Answer 9

integral (y(t)-x(t))dt from 0 to intersection
thats the integral you need.

the link shows the graphs, think about what is meant by enclosed and it should make sense.
http://www.wolframalpha.com/input/?i=intersection+x%5E2-2x+and+x%5E%281%2F2%29

Answer 10

ook no no no i think you have it wrong. i plug in the two parametric equations into my calc and the graph is WAY different. lemme attach a file to help you understand what I mean

Answer 11

what you have is not parametric AT ALL

Answer 12

please note, that they intersect at the same rate.
plug in the linear intersection and you will see that is the time that it is crossing the y axis the 2nd time. A root is a root.
That said, my integral was wrong because I thought you were dealing with normal functions.

Answer 13

Nothing wrong with ^^that assertion. However this is to me thought of as one curve, not two. so you will see a graph like i have in my picture. so the integral is what then? what i orginially have? because that would yeild a negative answer

Answer 14

lemme look it up real quick, I dont entirely remember how to do these parametrically, meantime, here's a java applet parametric grapher:
http://webspace.ship.edu/msrenault/ggb/parametric_grapher.html

Answer 15

ok cool, reply when you can. thanks for all the help so far btw

Answer 16

You're correct that the general form is:
\[\int\limits_{t_1}^{t_2}ydx\]
That said, you still need the intersection points (because they are the same as when the graph crosses the y axis, which is what you want).
\[\int\limits_{0}^{1/4 (1+ \sqrt(5))^2}(\sqrt t*(t^2-2t))dt\]

Answer 17

actually, I messed that up, apparently I cant pay attention tonight, lol
\[\int\sqrt t d(t^2-2t)\]
\[=\int\sqrt t *(2t-2)dt\]
left the limits off for brevity

Answer 18

You can probably just use 0 and 2 as your limits... it's close enough that your professor wont know the difference... its a 0.671 or something difference.

Answer 19

can i ask where the two comes from sorry im lost ahah im trying !

Answer 20

which 2? in the limits?
the above noted intersection is about 2.617
so, if you round to 2, you'll probably be fine... instead of carrying it as that radical and junk i put above

Answer 21

http://www.mathwords.com/a/area_parametric.htm

Answer 22

nice! cannot thank you enough for your time and help agreene!