Joeyboiiiiii
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2. The coordinates for the centroid of a lamana is calculated using the summation formulas given below:
. . . . .\(\displaystyle \bar{x}\, =\, \dfrac{\sum_{x=a}^{x=b}x\delta A}{\sum_a^b \delta A}\)
. . . . .\(\displaystyle \bar{y}\, =\, \dfrac{\sum_{y=a}^{y=b}y\delta A}{\sum_a^b \delta A}\)
a. Derive, from first principles, formulas for \(\displaystyle \bar{x}\) and \(\displaystyle \bar{y}\) involving integrals.
b. A Guillotine blade in the form of a lamina is shown in the graphic. Use your derived formulas for \(\displaystyle \bar{x}\) and \(\displaystyle \bar{y}\) to calculate its centroid. (The curve has equation \(\displaystyle y\, =\, x\, (3\, -\, x).\) The left-hand end is at \(\displaystyle x\, =\, 0.5.\))
This question has has me stuck for about three weeks now
. . . . .\(\displaystyle \bar{x}\, =\, \dfrac{\sum_{x=a}^{x=b}x\delta A}{\sum_a^b \delta A}\)
. . . . .\(\displaystyle \bar{y}\, =\, \dfrac{\sum_{y=a}^{y=b}y\delta A}{\sum_a^b \delta A}\)
a. Derive, from first principles, formulas for \(\displaystyle \bar{x}\) and \(\displaystyle \bar{y}\) involving integrals.
b. A Guillotine blade in the form of a lamina is shown in the graphic. Use your derived formulas for \(\displaystyle \bar{x}\) and \(\displaystyle \bar{y}\) to calculate its centroid. (The curve has equation \(\displaystyle y\, =\, x\, (3\, -\, x).\) The left-hand end is at \(\displaystyle x\, =\, 0.5.\))

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