Question

1. The unbounded plan ar region bet ween the z-axis and the curve yis revolved about the r2 1 T-axis (a) Find the volume of the resulting solid of revolution b) Does the solid have faite surface area? Justify sour answer carefilly

Answer 1

we are given

$y=\frac{1}{x^2+1}$

y=0

(a)

we can draw graph

2 z) 2 0 2

we can set up integral for volume

$V=\int _{-\infty \:}^{\infty \:}\pi \left(\frac{1}{x^2+1}\right)^2dx$

Firstly, we will solve integral

$\int \:\pi \left(\frac{1}{x^2+1}\right)^2dx$

2 1)2

π. sin (2 arctan (z)) ) arctan (z) +

now, we can plug bounds

and we get

부 (-?)

$V=\frac{\pi ^2}{2}$...............Answer

(b)

$y=\frac{1}{x^2+1}$

Firstly, we will find derivative

2 dr _ (x2 + 1)2

now, we can find ds

$ds=\sqrt{1+(-\frac{2x}{(x^2+1)^2})^2}dx$

Since, curve goes to infinity

so, length of curve will be infinite

and surface area will also be infinity

So,

Answer is NO