Decide the Center of Mass for Rotational Body K

[Pir]

Homework Statement

The curve y = sinx\sqrt{3cosx}, <br /> 0\leq x\leq \pi /2 rotates around the x-axis and creates a homogenous rotational body K.

a) Decide the volume of K.

b) Decide the center of mass for K.

(The x-coordinate of the center of mass is X_{T} = \frac{1}{m}\int_{K}^{} x dm, where m is the mass of of K.)

Homework Equations

X_{T} = \frac{1}{m}\int_{K}^{} x dm

The Attempt at a Solution

I have decided a) and I got the volume to be ∏ volume units. I need help with b).

I try to use the formula and I get this:

X_{T} = \frac{1}{\pi}\int_{K}^{} 3\pi x sin^2xcosx dx

But I need help how to solve this integral (if it's correct?). Please help me with this, how do I solve this integral?

 

[Dick]

Homework Statement

Homework Equations

X_{T} = \frac{1}{m}\int_{K}^{} x dm

The Attempt at a Solution

I have decided a) and I got the volume to be ∏ volume units. I need help with b).

I try to use the formula and I get this:

X_{T} = \frac{1}{\pi}\int_{K}^{} 3\pi x sin^2xcosx dx

But I need help how to solve this integral (if it's correct?). Please help me with this, how do I solve this integral?

Did you try integration by parts?

 

[Pir]

Is that formula correct to begin with?

I tried integrating by parts but I couldn't solve it. Please help by writing how to do it. I have a test in two days and I need to know how to solve this before then and it takes forever if I have to ask for one step at a time when there might be several questions.

 

[Dick]

Is that formula correct to begin with?

I tried integrating by parts but I couldn't solve it. Please help by writing how to do it. I have a test in two days and I need to know how to solve this before then and it takes forever if I have to ask for one step at a time when there might be several questions.

Yes, it looks correct. Try parts using u=x dv=sin(x)^2*cos(x)dx.

 

[Pir]

Yes, it looks correct. Try parts using u=x dv=sin(x)^2*cos(x)dx.

I don't understand, please write how to do it. You mean substitute x with u? That doesn't really change anything, there are still three factors.

 

[Dick]

I don't understand, please write how to do it. You mean substitute x with u? That doesn't really change anything, there are still three factors.

No, I mean do integration by parts with those as the parts. $\int u dv=uv-\int v du$. That's integration by parts.

 

[Pir]

wrong

 

[Pir]

What happens with the pi?

And why isn't the x integrated with the rest? Shouldn't it be 1/2 x^2?

 

[Dick]

What happens with the pi?

And why isn't the x integrated with the rest? Shouldn't it be 1/2 x^2?

You can just factor out the constants like 3 and pi. And it doesn't sound like you done integration by parts before. There are some examples here: http://en.wikipedia.org/wiki/Integration_by_parts Start from u=x and dv=sin(x)^2*cos(x)dx. Try to figure out what v is by integrating dv.

 

[Pir]

Yes the pi can be factored out but DID YOU NOTICE THAT THERE'S ALSO A 1/PI, DICK? What happens with that?

 

[Dick]

Yes the pi can be factored out but DID YOU NOTICE THAT THERE'S ALSO A 1/PI, DICK? What happens with that?

I told you. Keep track of the constants separately. I'm trying to show you how to integrate x*sin(x)^2*cos(x). The constants are the easy part.