I need to grade some work, and there's an integral to calcaulte:(adsbygoogle = window.adsbygoogle || []).push({});

compute the area of y^2+z^2=2ax where x is between a and y^2/a.

I tried using the coordinates:

[itex]y=\sqrt{2as}\cos(t) \ z=\sqrt{2as}\sin(t) \ x=s[/itex]

But I don't see how to find the limits of integration with respect to s and t.

Thanks.

The higher you get in maths, the less you remember what you learned in earlier years. :-(

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# A simple calculation that got me stumped.

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