# Find the moment about the y-axis

1. Nov 3, 2009

### xstetsonx

Given the region defined by y=6(x)^.5-4 and y=6x^2-4 with density p, Find the moment about the y-axis

I found the upper bound but coulden't find the lower bound.

and i just want to make sure i set up the equation right

my=p$$\int$$ (x)((6(x)^.5-4)-(6x^2-4))dx

I used the shell method in case i confuse anybody

2. Nov 3, 2009

### Staff: Mentor

Actually, you used thin vertical strips, not shells, which are used when a region is rotated around an axis. By "lower bound" I assume you mean the bottom limit of integration. Set 6sqrt(x) - 4 = 6x2 - 4, and solve for x. You should find two values of x. Those are your limits of integration.

Other than that, your integral looks fine.

3. Nov 3, 2009

### xstetsonx

so if i set those two equation up i will get
6x^.5(-x^(3/2)+1)
which will give me 0 and 1 right?

4. Nov 3, 2009

### Staff: Mentor

You should have an equation.
6x^.5(-x^(3/2)+1) is not an equation.

5. Nov 3, 2009

### xstetsonx

i just set it to 0
6x^.5(-x^(3/2)+1)=0
6x^.5=0
x=0
-x^(3/2)+1=0
x=1

6. Nov 3, 2009

### Staff: Mentor

OK, that's an equation. And yes, the x values at the intersection points are at x = 0 and x = 1. BTW, there are two more values of x, but they are complex.

7. Nov 3, 2009

### xstetsonx

do they involve imaginary number?

+ and - i????

Last edited: Nov 3, 2009
8. Nov 3, 2009

### Staff: Mentor

Yes, but since you're interested only in real solutions, you can ignore them.

What I did was solve 6sqrt(x) - 4 = 6x^2 -4 ==> 6sqrt(x) = 6x^2 ==> sqrt(x) = x^2. I squared both sides to get x = x^4 ==> x(x^3 - 1) = 0. The left side can be factored.
x(x - 1)(x^2 + x + 1) = 0, so x = 0, x = 1, or x^2 + x + 1 = 0. The quadratic has complex solutions.

9. Nov 3, 2009

### xstetsonx

YES can't believe i still remember stuff from cal 1 haha thanks so much u r AWESOME!!!!!

10. Nov 3, 2009

### Staff: Mentor

Thanks much! I appreciate the feedback.