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How do you find the centroid of this?

  • #1

Homework Statement


Find the centroid of the shape formed by the equation y2=x3-x4, the x-axis, and the y-axis.

Homework Equations


A=∫f(x)dx
Mx=∫(1/2)[f(x)]2dx
My=∫x[f(x)]dx

The Attempt at a Solution


I'm stuck on the integral.
I attempted u-substitution and got du=(1/2)(x3-x4)-1/2dx; "parts," trigonometric substitution/identities, and partial fractions don't seem to apply.
 

Answers and Replies

  • #2
SammyS
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Science Advisor
Homework Helper
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Homework Statement


Find the centroid of the shape formed by the equation y2=x3-x4, the x-axis, and the y-axis.

Homework Equations


A=∫f(x)dx
Mx=∫(1/2)[f(x)]2dx
My=∫x[f(x)]dx

The Attempt at a Solution


I'm stuck on the integral.
I attempted u-substitution and got du=(1/2)(x3-x4)-1/2dx; "parts," trigonometric substitution/identities, and partial fractions don't seem to apply.
Try polar coordinates.
 
  • #3
Try polar coordinates.
I haven't worked in those for a while (I'm not sure I remember how to use them); but I'm fairly certain my teacher wants me to stick with x-y coordinates.
 
  • #4
SteamKing
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Science Advisor
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I haven't worked in those for a while (I'm not sure I remember how to use them); but I'm fairly certain my teacher wants me to stick with x-y coordinates.
You've already tried trig substitutions. Think of polar coordinates as a form of trig substitution.

Besides, what will your teacher prefer? That you kept working with cartesian coordinates and didn't solve the problem, or you converted to polar coordinates and got an answer? :wink:
 
  • #5
vela
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Homework Statement


Find the centroid of the shape formed by the equation y2=x3-x4, the x-axis, and the y-axis.

Homework Equations


A=∫f(x)dx
Mx=∫(1/2)[f(x)]2dx
My=∫x[f(x)]dx
You swapped ##M_x## and ##M_y##.

The Attempt at a Solution


I'm stuck on the integral.
I attempted u-substitution and got du=(1/2)(x3-x4)-1/2dx; "parts," trigonometric substitution/identities, and partial fractions don't seem to apply.
Try starting like this:
$$\int x\sqrt{x^3-x^4}\,dx = \int x^2\sqrt{x-x^2}\,dx = \int x^2\sqrt{\frac 14-\left(x-\frac 12\right)^2}\,dx.$$ The last step comes from completing the square. Then try a few more substitutions and see if you get anywhere.
 

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