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.

Answer 1

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

Answer 2

or is it cos^2?

Answer 3

regardless; its the volume of:

pi [S] [1/(1+x^2)]^2 dx ; [0,3]

Answer 4

x = tan

dx = sec^2 dt

sec^2 dt
------.....really? t? lol
sec^2

Answer 5

nah...

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

Answer 6

cos^2 = 1/2 + cos(2t)/2

t/2 + sin(2t)/4 is our integration

Answer 7

t = tan^-1(x) so...

tan^-1(x)/2 + cos(2(tan^-1(x)))/2

Answer 8

35.78 -0.4 = 31.78....or so

Answer 9

that wa typoed lol...ack!!

Answer 10

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

Answer 11

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 :)