Finding Surface area of a Parametric Curve

[trap]

Can someone please help me with this question?

x = 1-sint, y = 2+cost, rotate about y = 2

Find the surface area of the parametric curve.

I don't know how to do it with y=2, I only know how if the question askes for rotating about the x-axis.
The answer to the question is 2(pi)^2.

 

[amcavoy]

Can someone please help me with this question?

x = 1-sint, y = 2+cost, rotate about y = 2

Find the surface area of the parametric curve.

I don't know how to do it with y=2, I only know how if the question askes for rotating about the x-axis.
The answer to the question is 2(pi)^2.

Since you are rotating about y=2, that makes each of your y-values 2 less, so the equations become:

x = 1 - sin(t)
y = cos(t)

Now I think you can do the rest:

$$\text{SA}=2\pi\int_{0}^{2\pi}y\,\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt$$

 

[trap]

I forgot to mention that the interval is t between (-pi/2, pi/2), how do i get the answer 2pi^2 with

$$\text{SA}=2\pi\int_{-\pi/2}^{\pi/2}y\,\sqrt{\left(\frac{ dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt$$ ??

so i have... dx = -cost
dy = -sint

here's what I've done..but couldn't get the answer

$$\text{SA}=2\pi\int_{-\pi/2}^{\pi/2}y\,\sqrt{(sint)^2+(cost)^2}\, dt$$

$$=2\pi\int_{-\pi/2}^{\pi/2}y\,\sqrt{1}\, dt$$

$$=2\pi\int_{-\pi/2}^{\pi/2}cost\, dt$$

$$=2\pi\sin(\pi/2) - 2\pi\sin(-\pi/2)$$

$$=4\pi$$

 

[trap]

but i got the formula of surface area from my textbook, which is,

$$\text{SA}=2\pi\int_{}^{}y\,\sqrt{\left(\frac{ dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt$$

 

[amcavoy]

but i got the formula of surface area from my textbook, which is,

$$\text{SA}=2\pi\int_{}^{}y\,\sqrt{\left(\frac{ dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt$$

Yes, right. I fixed it sorry about that. Anyways, using this you should come up with your answer (I myself am coming up with -4π). Are you sure it's 2π²?

The only way I get 2π² is by the following:

$$2\pi\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\cos^2{t}\,dt$$

...but you clearly don't have a cos²(t)...

 

[trap]

yes..the answer is supposed to be 2pi^2, unless it is a typo in the textbook

 

[amcavoy]

yes..the answer is supposed to be 2pi^2, unless it is a typo in the textbook

It happens but I doubt that. Does anyone have any ideas why the work above isn't giving the correct answer?

 

[myininaya]

i think you were suppose to multiplied by x(t) not y(t)