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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rakesh05 | High School Teacher | (Level 1) Assistant Educator

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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.

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

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pramodpandey | College Teacher | (Level 3) Valedictorian

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 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






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oldnick | (Level 1) Valedictorian

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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`

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oldnick | (Level 1) Valedictorian

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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` 



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oldnick | (Level 1) Valedictorian

Posted on

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


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