Volume with Parametric Equations

**Quasar** · February 23rd 2008, 10:07 AM

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:

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

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

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

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 $(1,\frac{5}{2})$ and $(1,1)$ to the y-axis; and one for the parametric curve.

I have been trying to use the shell method with the shell radius being $y$ , and $y=\frac{3}{2}t^2+1$ 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.

**Quasar** · February 23rd 2008, 02:53 PM

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

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

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

**galactus** · February 23rd 2008, 04:15 PM

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

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

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

I hope this helps. Let me know.

**Quasar** · February 23rd 2008, 04:25 PM

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.

**CubiX** · October 19th 2009, 09:26 AM

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

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

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:
$V_x=\pi\int_{t1}^{t2}y(t)^2x'(t)dt$ .

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

**CubiX** · October 21st 2009, 04:35 AM

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

Thx

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