Thread: Volume with Parametric Equations

I've been banging my head against this for a couple days now. The first question is to find the area and the second question (where I am really getting stuck) is to take that area and find a volume. This question is from school but I only have one example in my notes and as much as I look at it...it still makes no sense to me.

The Parametric Equations are as follows:

$\displaystyle x=t^3-3t+1$ and $\displaystyle y=\frac{3}{2}t^2+1$

Question 1: Find the area between the curve and the y-axis from $\displaystyle t=-\sqrt{3}$ to $\displaystyle t=0$

So I set this up as Area = $\displaystyle \int_{t1}^{t2} x dy$
When $\displaystyle dy=3tdt$
Area = $\displaystyle \int_{-\sqrt{3}}^{0} (t^3-3t+1)(3t)dt$ and I worked this out to be Area = $\displaystyle \frac{18\sqrt{3}}{5}+\frac{9}{2}$

Question 2: What is the volume if the area (in question 1) is revolved about the x-axis?

This is where I am stuck. The first question may be wrong but atleast I can begin to work it. I don't know where to start setting up the integral to find the volume. I am thinking that the set up will have two integrals: One for the two horizontal lines that go from points $\displaystyle (1,\frac{5}{2})$ and $\displaystyle (1,1)$ to the y-axis; and one for the parametric curve.

I have been trying to use the shell method with the shell radius being $\displaystyle y$, and $\displaystyle y=\frac{3}{2}t^2+1$ but I don't feel comfortable with anything I have set up.

I would greatly appreciate any help you can offer.
Thank you in advance.

Isn't that formula for surface area? I am trying to find the volume.

Now I am thinking $\displaystyle \pi\int_a^b (y\frac{dx}{dt}-x\frac{dy}{dt})dt$

Does that sound right?
I have been all over the internet and through all the books that I have and nobody seems to talk about volume when using parametric equations. I've been working on this problem for so long now, I'm starting to feel like this little guy

Come to think of it, I beleive the formula for a revolution about the x-axis in parametric is given by

$\displaystyle {\pi}\int_{a}^{b}[y(t)]^{2}\cdot{(x'(t))}]dt$

In your case, $\displaystyle {\pi}\int_{-\sqrt{3}}^{\sqrt{3}}\frac{3(t^{2}-1)(3t^{2}+2)^{2}}{4}dt=\int_{-\sqrt{3}}^{\sqrt{3}}[\frac{27t^{6}}{4}+\frac{9t^{4}}{4}-6t^{2}-3]=\frac{1476{\pi}\sqrt{3}}{35}\approx{229.47}$

I hope this helps. Let me know.

Well, if that is correct, then at leaset that means I was getting warmer!

Can you help me think through the process of getting the integral set up? I have a fair understanding of more basic volume problems, but the parametric equations confuse me. I feel like I almost understand.......and then I fall on my face.

I'm working on a similar problem. I have a parametric equation like so:
$\displaystyle x(t) = 2t^3 - \frac{3t}{2} + 2$
$\displaystyle y(t) = 3t^2 + 2$.

I first need to find the value of the parameter t, where the curve intersects itself. I have found this to be values $\displaystyle -\frac{\sqrt{3}}{2}$ and $\displaystyle \frac{\sqrt{3}}{2}$.

Now I need to find the volume of the revolution of the curve with the above lower and upper limit, both over the x-axis and y-axis.

This is supposed to be the formula for the x revolution:
$\displaystyle V_x=\pi\int_{t1}^{t2}y(t)^2x'(t)dt$.

However I'm not getting the correct result. Am I using the formula incorrectly?
$\displaystyle \pi\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}(3t^2 + 2)^2(2t^3 - \frac{3t}{2} + 2)'dt$

Never mind. It turns out that I got the correct answer after all.

Thx