Find the moment about the y-axis

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Homework Help Overview

The problem involves finding the moment about the y-axis for a region defined by the curves y=6(x)^0.5-4 and y=6x^2-4, with a given density p. Participants are discussing the setup of the integral and the bounds of integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the integral for the moment about the y-axis and the identification of bounds of integration. There is some confusion regarding the method used (shell method vs. vertical strips) and the correct interpretation of the lower bound.

Discussion Status

The discussion is ongoing, with participants clarifying the setup of the equations and the limits of integration. Some have confirmed the intersection points at x = 0 and x = 1, while others have noted the existence of complex solutions, which are not relevant to the current problem.

Contextual Notes

Participants are working under the constraints of a homework assignment, focusing on real solutions and the correct interpretation of the problem setup. There is an emphasis on ensuring that the equations are properly defined and understood.

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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[tex]\int[/tex] (x)((6(x)^.5-4)-(6x^2-4))dx

I used the shell method in case i confuse anybody
 
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xstetsonx said:
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[tex]\int[/tex] (x)((6(x)^.5-4)-(6x^2-4))dx

I used the shell method in case i confuse anybody

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.
 
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?
 
You should have an equation.
6x^.5(-x^(3/2)+1) is not an equation.
 
Mark44 said:
You should have an equation.
6x^.5(-x^(3/2)+1) is not an equation.

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
 
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.
 
do they involve imaginary number?

+ and - i?
 
Last edited:
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.
 
YES can't believe i still remember stuff from cal 1 haha thanks so much u r AWESOME!
 
  • #10
Thanks much! I appreciate the feedback.
 

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