Question

headless snake on the y axis is given by : (z-siny)^2+x^2=a^2 where 0 14 years ago

Answer 1

What do you think about these intervals for integration?\[(z-\sin y)^2\le a^2\to \sin y-a\le z\le\sin y+a\]\[ x\le a^2\to-a\le x\le a\]\[0\le y\le 8\pi\]I'm not sure, but that's all I can figure...

Answer 2

you did right my firend .. but i don't understand why x is between-a and a

Answer 3

the smallest that absolute value of |z-siny| can be is 0, and when it is we have\[x^2=a^2\to x=\pm a\] so|a| is the max value that x can take on

Answer 4

and about z, according to my solution is going like this: from -sqrt(a^2-x^2)+siny to the same but positive

Answer 5

yeah I had that at first, but again if you look at the region in the xz-plane under the region, it is a circlethe smallest |x| can be is 0, which means the max value that the other component of the circle can be is\[(z-\sin y)^2=a^2\to z-\sin y=\pm a\]you don't need to keep x because the only constraints on the max and min values of z-siny come from the radius of the circle: a

Answer 6

the region under the solid is a circle*

Answer 7

ohh.. nice ... didin't think about it !!!

Answer 8

I hope that made sense...
welcome :D