The objective is to find the volume of the solid generated by revolving the curve y=a^3/(a^2+x^2) about its asymptote.

ormaybesaladqh

ormaybesaladqh

Answered question

2022-10-21

Finding the volume of the solid generated by revolving the given curve.
The objective is to find the volume of the solid generated by revolving the curve y = a 3 a 2 + x 2 about its asymptote.
Observing the given function yields that y 0, hence y = 0 is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about x a x i s is gives as V = 2 π 0 ( f ( x ) 2 0 ) d x.
Which gives: 2 π 0 a 6 ( a 2 + x 2 ) 2 d x
Now, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form 1 x 4 + 2 x 2 + 1 for a = 1, which is transformed into 1 x 2 x 2 + 2 + 1 x 2 = 2 [ 1 + 1 x 2 x 2 + 2 + 1 x 2 ( 1 1 x 2 ) x 2 + 2 + 1 x 2 ] which can be integrated, but this integral is tending to infinity.

Answer & Explanation

Kenley Rasmussen

Kenley Rasmussen

Beginner2022-10-22Added 13 answers

Step 1
Neither of your two integrals in the last step are divergent. The first one is
I 1 = 0 1 + 1 x 2 x 2 + 2 + 1 x 2 d x = 0 x 2 + 1 x 4 + 2 x 2 + 1 d x = 0 d x x 2 + 1 = [ arctan x ] 0 = π 2 .
Step 2
So that's nice and convergent. The other one is a bit trickier, but we can find it with a quick jaunt into the complex plane:
1 1 x 2 x 2 + 2 + 1 x 2 d x = x 2 1 x 4 + 2 x 2 + 1 d x = 1 2 [ 1 ( x + i ) 2 + 1 ( x i ) 2 ] = 1 2 [ 1 x + i + 1 x i ] = 1 2 2 x x 2 + 1 = x x 2 + 1
and so 0 1 1 x 2 x 2 + 2 + 1 x 2 d x = [ x x 2 + 1 ] 0 = 0..
Raiden Barr

Raiden Barr

Beginner2022-10-23Added 7 answers

Step 1
This seems like a trig substitution problem to me, albeit a not-so-straightforward one.
Let x = a tan θ. Then d x = a sec 2 θ d θ, and a 2 + x 2 = a 2 sec 2 θ. Since θ = tan 1 ( x a ) , θ = 0 when x = 0, and θ π 2 as x . Therefore
2 π 0 a 6 ( a 2 + x 2 ) 2 d θ = 2 π 0 π / 2 a 6 a 4 sec 4 θ a sec 2 θ d θ = 2 π a 3 0 π / 2 cos 2 θ d θ = π a 3 0 π / 2 ( 1 + cos 2 θ ) d θ = π a 3 [ θ + 1 2 sin 2 θ ] 0 π / 2 = π 2 a 3 2
Step 2
I can't quite figure out how you intended to integrate
1 + 1 x 2 x 2 + 2 + 1 x 2 d x, but it could be that you were trying to represent a convergent improper integral as the difference of two divergent improper integrals. That leads to trouble, since is an indeterminate limit form.

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