Charge over a triangular region

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SUMMARY

The total charge calculation over a triangular region defined by vertices (0,0), (2,2), and (4,0) involves the volume charge density \(\rho_{v} = 6xy \, C/m^{2}\). The integral setup for the charge \(Q\) is given by \(Q = \int_{s} \rho_{v} ds\). The correct integration limits for \(x\) should be from 0 to 2, leading to the correct total charge of 32C, as opposed to the incorrect calculation of -128C due to misapplied limits.

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Homework Statement


-Given that [tex]\rho_{v}[/tex]=6xy C/[tex]m^{2}[/tex], Calculate the total charge on the triangular region.

-the verticies of the triangle are: (0,0), (2,2) and (4,0)


Homework Equations



[tex]Q=\int_{s}\rho_{v}ds[/tex]

The Attempt at a Solution



[tex]=\int^{4}_{0}\int^{-x+4}_{x}6xy dydx[/tex]
[tex]=\int^{4}_{0}48x - 24x^{2} dx[/tex]
[tex]=24x^{2}-8x^{3}|^{4}_{0}[/tex]
[tex]=-128C[/tex]

According to the textbook the correct answer is 32C

any help would be appreciated!
Thanks,
 
Physics news on Phys.org
You got your integration limits mixed up. The maximum value for x is 2 m. Draw a picture and you will see why.
 

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