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.

Question

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.

amistre64 · Answer

sec(t)^2 is what that amounts to ... right?

amistre64 · Answer

or is it cos^2?

amistre64 · Answer

regardless; its the volume of:

pi [S] [1/(1+x^2)]^2 dx ; [0,3]

amistre64 · Answer

x = tan

dx = sec^2 dt

sec^2 dt
------.....really? t? lol
sec^2

amistre64 · Answer

nah...

 sec^2 dt
------     =  cos^2 dt  thats it
sec^4

amistre64 · Answer

cos^2 = 1/2 + cos(2t)/2

t/2 + sin(2t)/4  is our integration

amistre64 · Answer

t = tan^-1(x)  so...

tan^-1(x)/2  + cos(2(tan^-1(x)))/2

amistre64 · Answer

35.78 -0.4 = 31.78....or so

amistre64 · Answer

that wa typoed lol...ack!!

amistre64 · Answer

tan^-1(x)/2  + sin(2(tan^-1(x))/4

amistre64 · Answer

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 :)