Question

Find the area enclosed by the x-axis and the curve x=t^3, y=1-t^6.

Answer 1

Since it is just the x-axis... i do not need to subtract anything? just integrate y(dx/dt)dt right?

Answer 2

this might help http://www.mathwords.com/a/area_under_a_curve.htm

Answer 3

did you get an answer?

Answer 4

\[t^3=x\]
\[t=\sqrt[3]{x}\]
Replace t in the equation for y to get the equation in terms of x
\[y=1-(\sqrt[3]{x})^6\]
Expand the exponent of 6 to each factor in the expression
y=1-x^2
Multiply -1 by the X6^2 inside the parentheses
\[y=1-x^2\]

Not positive if this is right do you have choices?

Answer 5

no choices. and I need to use calculus on it. The answer should be a number. to find area I integrate from a to b ... with in this case is 0 to 1 of y (dx/dt) dt. I got 2/3 sq units.. do not know if that is correct.

Answer 6

@radar can you help?

Answer 7

thanks for helping me by the way.

Answer 8

sorry wish I could help your in Geometry

Answer 9

hehe Multivariate Calculus.

Answer 10

lol i better not help you :) i might mess you up. Im in geometry just getting into trig

Answer 11

fun times! Trig is cool stuff.

Answer 12

my brain is fried from last semester but my teacher texted and did say I got a B so ill take that. :)

Answer 13

ok ttyl I hope someone comes and helps you

Answer 14

thanks for helping @radar

Answer 15

If I just had.. "Need help finding area" as my question title.. more people would come in.....

Answer 16

@thomaster @Mertsj maybe you can help? I got 2/3 sq units as an answer. Thanks in advance!

Answer 17

y = -x^2 +1, is it not a downward parabola with the vertex is (0,1)?
so, the area under the curve is calculated by int from -1 to 1 of this curve?

Answer 18

Sorry I couldn't help, I keep telling myself to review calculus, but never seem to get to it.

Answer 19

so I need to integrate from -1 to 1?

Answer 20

ahh since this is an ellipse?

Answer 21

you have parametric equations, and bink give you the equation above, To me, it's good enough to calculate. and Yes, take integral from -1 to 1 of -x^2 +1
parabola, not ellipse

Answer 22

the answer says find the area enclosed by the ellipse.. is why i said that

Answer 23

I am trying to use the formula A=int (a,b) y(dx/dt) dt

Answer 24

do I not just start by calculating dx/dt?

Answer 25

When I graph this parametric curve.. it does not cross the y axis?

Answer 26

I don't see any reason to go to multivariate problem.
y = -x^2 +1, clear enough to have a curve,; clear enough to take integral

Answer 27

why not? y= 1- t^6, t =0 , y = 1
x = t^3 , t=0 x =0
it perfectly matches with y = -x^2 +1
you see, x =0 , y =1 is vertex of parabola y = -x^2 +1

Answer 28

so the area is 4/3?

Answer 29

I just plugged it into my graphing utility as a parametric curve.. and it only graphed in for positive values of x. I reset the domain.... and it still would not graph negative values of x.

Answer 30

I don't know, I don't calculate it, just give you my opinion

Answer 31

I really need to use multivariate procedures on this since that is my class.

Answer 32

oops not an ellipse, you are right.. just a curve.

Answer 33

@Loser66 thank you for your help! I understand the limits of integration now... the problem worked out the same whether I used the multivariate method or the method you showed. Sorry, I was reading 2 problems at once.

Answer 34

how genius you are. I read just 1problem/time. And when I got problem, I want to cry with this only. YOU, 2problems/ time. ha!!!! jealous

Answer 35

only 2 problems because I had to continue with my HW until help came.. which was a long time. It seems like the higher I go in math the more time it takes here to get help. Now trying to tackle arc length... long problems. Thanks again for your help!

Answer 36

hihihi, gooooooodddd luck.