# 1) What are the areas bounded by the lines and the x axis? In each case x>0 a) y = 3/xb) y=12xc) y=(1/12)x2)Find the volume V of the solid obtained by rotating the bounded area about the x...

1) What are the areas bounded by the lines and the x axis? In each case x>0

a) y = 3/x

b) y=12x

c) y=(1/12)x

2)

Find the volume V of the solid obtained by rotating the bounded area about the x axis

y=4-(1/2)x, y > 0, 2< x <3

*print*Print*list*Cite

### 1 Answer

1) a) `y = 3/x` , `x>0`

The area under the graph is given by

`int_0^infty 3/x dx = 3log(x)|_0^infty = 3[lim_(x-> infty)log(x) - lim_(x->0)log(x)] `

`= lim_(x-> infty) x`

b) `y = 12x`, `x>0`

The are under the graph is given by

`int_0^infty12x dx = 6x^2|_0^infty = lim_(x-> infty)6x^2 - 0 = lim_(x->infty) x`

c) `y = 1/(12x)` , `x>0`

The area under the graph is given by

`int_0^infty 1/12x dx = 1/24x^2|_0^infty = 1/24lim_(x->infty) x^2 - 0 ``= lim_(x->infty)x`

2) We want the volume V of the line

`y = 4 - 1/2x`, `2<x<3`

rotated about the x axis. We obtain this by integrating over discs of radius `y` from `x =` 2 to 3. The area of each disc is then `piy^2`.

`V = int_2^3 pi(4-1/2x)^2 dx = int_2^3 pi(16 - 4x + 1/4x^2)dx`

`= pi(16x - 2x^2 + 1/12x^3)_2^3 = pi(48-18+27/12 - 32+8 - 8/12)`

`= pi(6 + 19/12) = (91/12)pi`

The volume V is called a *solid of revolution*.

**1) The areas all tend towards infinity**

**2) V = (91/12)pi**