On this exercise (compute a center of mass using weighted average integral) : http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-c-average-value-probability-and-numerical-integration/session-61-integrals-and-weighted-averages/MIT18_01SCF10_ex61sol.pdf

I don't understand how it is possible that the total area computed using dx IS NOT equal to the same area computed using dy? I understand the calculation, but can't figure out why conceptually, one way of computing the area gives 20/3 and the other 32/3...

Question

On this exercise (compute a center of mass using weighted average integral) : http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-3-the-definite-integral-and-its-applications/part-c-average-value-probability-and-numerical-integration/session-61-integrals-and-weighted-averages/MIT18_01SCF10_ex61sol.pdf

I don't understand how it is possible that the total area computed using dx IS NOT equal to the same area computed using dy? I understand the calculation, but can't figure out why conceptually, one way of computing the area gives 20/3 and the other 32/3...

phi · Answer

they show the area integrating over x  is $ 10 \frac{2}{3} = \frac{32}{3} $
this matches the area integrating over y.

anonymous · Answer

Thank you for answering phi! Hm, there must be something I am missing here... How can 10*(2/3) = 32/3? Isn't 10*(2/3) = 20/3?

phi · Answer

$$ 10 \frac{2}{3} $$ means 10 plus 2/3
i.e. in decimals, 10.6666...

phi · Answer

read it as "ten and two-thirds"

anonymous · Answer

Ahhhhhhhhh ok! I would have never guessed the correct interpretation. oO Thank you very much! :)

phi · Answer

if you evaluate the integral by hand
$$ \frac{3}{2} x^2 - \frac{1}{4} x^4 + \frac{2}{3} x^3 \bigg|_{-1}^3= $$
you will get the correct number, and see it is not 20/3 = 6 2/3