I need help solving some problems including area bounded by a region 1) y = 3/xy=12xy=(1/12)xx>0
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
So, Area =(1/2)6(1/2)+`int_(1/2)^6 (3/x)dx`
=`(3/2)+3[logx]` between 1/2 to 6.
We wish to find area bounded by region ,green ,red and blue curves. Green and red intersect at x=1/2 ,Red and blue intersect at x=6 , and region satisfy all conditions is in first quadrant.
Thus area bounded region by curves
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`
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`
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`
The figure you have to rotate is thatammong red blue and green line