Areas dealing with parametric curves

[paulxzt]

Estimate the area of the region enclosed by the loop of the curve x=t^3-12t and y=3t^2+2t+5.

Can someone please help me with this..

I solved for t in the x equation to get t = + or - sqrt(12)
then the formula A = integral of b to a of g(t)*f'(t) dt
i set the bounds from -sqrt(12 to sqrt(12) and got 798.

im pretty sure i totally screwed it up, anyone please help

 

[Deleted member 4993]

Estimate the area of the region enclosed by the loop of the curve x=t^3-12t and y=3t^2+2x+5.

Are your equations correct?

 

[paulxzt]

Estimate the area of the region enclosed by the loop of the curve x=t^3-12t and y=3t^2+2x+5.

Are your equations correct?

Oops.. obviously a typo.. fixed it to t...

 

[galactus]

Your limits of integration should be 0 to 2PI, shouldn't they?.

oriented clockwise:

\(\displaystyle \L\\\int_{0}^{2{\pi}}[y(t)x'(t)]dt\)

or

\(\displaystyle \L\\-\int_{0}^{2{\pi}}[x(t)y'(t)]dt\)

 

[paulxzt]

Your limits of integration should be 0 to 2PI, shouldn't they?.

oriented clockwise:

\(\displaystyle \L\\\int_{0}^{2{\pi}}[y(t)x'(t)]dt\)

or

\(\displaystyle \L\\-\int_{0}^{2{\pi}}[x(t)y'(t)]dt\)

why is it 0 to 2pi? does the curve make a circle?
if i integrate that, I get a huge number = 17,377.468..

 

[galactus]

I assume then you did not graph it. Yes, it goes in a loop. The problem statement says that.

 

[paulxzt]

I assume then you did not graph it. Yes, it goes in a loop. The problem statement says that.

is there a way to figure out the bounds without graphing? how do you graph the x = function on a calculator or by hand? also, the area comes out to be a huge number.