Find the volume of the solid obtained by rotating the region R bounded by the curves , y =1/(1+x^2), y= 0, x = 0 and x=3 about the x-axis.

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Find the volume of the solid obtained by rotating the region R bounded by the curves , y =1/(1+x^2), y= 0, x = 0 and x=3 about the x-axis.

Mathematics
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sec(t)^2 is what that amounts to ... right?
or is it cos^2?
1 Attachment
regardless; its the volume of: pi [S] [1/(1+x^2)]^2 dx ; [0,3]

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Other answers:

x = tan dx = sec^2 dt sec^2 dt ------.....really? t? lol sec^2
nah... sec^2 dt ------ = cos^2 dt thats it sec^4
cos^2 = 1/2 + cos(2t)/2 t/2 + sin(2t)/4 is our integration
t = tan^-1(x) so... tan^-1(x)/2 + cos(2(tan^-1(x)))/2
35.78 -0.4 = 31.78....or so
that wa typoed lol...ack!!
tan^-1(x)/2 + sin(2(tan^-1(x))/4
we might be easier to just convert the interval to match the "t" tan(t) = x when x = 0 t=0 when x=3 t= 71.56 so; 0/2 + sin(2(0))/4 = 0 71.56/2 + sin(2(71.56))/4 = 35.78 + 0.15 = 35.93....hopefully :)

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