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Yep first year. Its nothing fancy, just sec blows up close to pi/2, and does so fast enough for the integral to not exist. You can check this using the definition of an improper integral. If you are interested in getting a head start on the early uni stuff check out Spivak's...

It is not particularly hard to integrate this for negative n (we are just finding the primitive of an even power of sec), although between the two limits given here such an integral will not converge.

Not really, I cannot imagine something being much faster. I suppose if you take as assumed the formula for the area of a circle, your linear transformation method is just as quick. We don't even need to do any integration as the Jacobian determinant is obviously constant. Just multiply the area...

Green's theorem provides another fast way of finding this area.

Haha cheers, I know you have high standards in that regard. Yeah the series is just the real part of a complex geometric series, so it is very quick to evaluate without induction.

Actually, I did that work in a bit of a rush and made a minor mistake in my layout. We are assuming that the identity holds for n-1, and then proving that it holds for n. We are NOT assuming that it holds for n=k-1 and proving that it holds for n=k.

It will work just the same, the expressions involved will just be marginally messier. You could literally copy the same proof and replace every instance of 'k' with 'k+1'.

Apologies for the handwriting...

Will scan and upload a solution in a sec.

15\pi/2-9\sqrt{3}\, \, cm^2 Hopefully I didn't make a calculation error.

I guess no cookie for me either! Kudos on compiling these together for students :).

Will be attending most likely, sounds like an excellent idea :).

Don't be silly, you cannot cross a scalar with a vector :D.

Yep haha, its clever. What do you get if you cross a mountain climber with a mosquito?

Zero! :D.

I thought the second person was meant to be a physicist or an engineer? Logicians if anything are even more pedantic than mathematicians...

Would rep but need to spread. More exam questions should involve finding the area of moustaches.

Yep, it falls apart from those two formulae. Note that the axis of rotation (which is a vertical line) cannot be between a and b, the theorem of Pappus is not valid here as our solid of revolution self intersects.

The theorem of Pappus is a useful tool for quickly calculating volumes (and surface areas) of solids of revolution. It states that if we obtain a solid of revolution by rotating a region of area A about a line l, then the resultant volume will be Ad, where d is the distance traced out by the...