Alan

The sum of the currents out of the junction equals the sum of the currents into the junction.

(Kirchoff's current law)

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I looked for a way to get the sin and cos terms the "same" in some sense, and I could see that if we had integer multiples of some common angle we could expand the result to get both in terms of the one angle.  

There might well be a more straightforward way of doing the integral!

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Here's a picture of the vectors for the cross-product:

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19.  Use the fact that the stored energy is (1/2)*Q²/C  where capacitance  C = εA/d

 and ε is permittivity, A is area and d is distance between plates.

You can work out Q from the fact that you are given the number of Coulombs per unit area (specified in terms of microCoulombs per unit area)) and you know the radius of a plate so you can find its area. 

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See graph:

a ≈ ±38.816

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For the first integral:

This should be carefully checked as I tend to be a little cavalier with constants sometimes!

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So how about:

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Try using the Poisson distribution Melody.

The probability of exactly k defects/unit interval, given an average of λ defects/unit interval is given by:

$$p(k;\lambda)=\frac{\lambda^ke^{-\lambda}}{k!}$$

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t is common to both terms so you can write: t(9.75t + 1.8) = 0

This mean that either t = 0 or t = -1.8/9.75  

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