# I need help solving some problems including area bounded by a region 1) y = 3/xy=12xy=(1/12)xx>0

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The figure you have to rotate is thatammong red blue and green line

First you have to found area of the figure rotating:

intersections: `3/x=12x rArr` `4x^2=1 rArr` `x=1/2`

`1/12 x =3/x` `rArr x=6`

so `f(1/2)= 6` and `f(6)= 1/2`

so `A=6 xx 1/2 xx 1/2 + int _(1/2)^6 3/x dx - 6xx1/2xx1/2= ` `3int_(1/2)^6 dx/x=` `3log 12`

The second step is to find moments of figures:

`M_x= int_(1/2)^6 y xx 3/x dx=` `int_(1/2)^6 9/x^2 dx` `=9 int_(1/2)^6 1/x^2 dx=33/2`

`M_y=int_(1/2)^6 x xx3/x dx=11/2`

Third step is to find G coordinates by means equations:

`G_x= M_x/A= 11/(2log12)=5.0964`

`G_y=M_y/A=11/(6log12)=0.7379`

Fourth step isto find the distanceof the batricent G from the axis of rotation:

I.e using formula distance `d` by a point `P(x_0;y_0)` by a straight line of equation `ax+by+c=0`

`d=( ax_0+by_0+c)/sqrt(a^2+b^2)`

By Pappo - Guldino theorem: `V= 2 pi d A`

In the case `x>0` X axis: `d= G_y=11/(6log12)` and `V=11 pi`

Thus area bounded region by curves

=`int_0^(1/2)12xdx+int_(1/2)^6(3/x)dx-int_0^6(1/12)xdx`

`=(6x^2)_0^(1/2)+3(log(x))_(1/2)^6-(1/24)(x^2)_0^6`

`=6/4+3(log(6)-log(1/2))-36/24`

`=3log(12)`

Ans.

We want to find the area bounded by y=3/x, y=12x and y=(1/12)x.

Clearly the desired region has its upper boundary y=12x and the lower boundary y=(1/12)x.

Solving Y=3/x and y=12x we get x=1/2 (as x>0).

and y=3/x and y=(1/12)x, we get x=6.

We know that Area bounded by the curve y=f(x) and on the x-axis between x=a to x=b is given by

Area=`int_a^bydx`

So, Area =(1/2)6(1/2)+`int_(1/2)^6 (3/x)dx`

=`(3/2)+3[logx]` between 1/2 to 6.

=(3/2)+3(log6+log2)=3/2+3(log12)=1.5+7.4547=8.9547. Ans.