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.