Lesson Find the volume of a solid obtained by rotation of some plane shape about an axis

Find the volume of a solid obtained by rotation of some plane shape about an axis

Problem 1

I will assume that "a" is positive real number.


Notice that this given triangle is SPECIAL. It has vertical leg from (a,a), to (a,2a) 
and horizontal leg from (a,2a) to (2a,2a).  


So, the triangle itself is a right angled triangle. It is "the upper half of the square"   a <= x, y <= 2a,
cut in two halves by its diagonal y = x.  It facilitates thinking and calculations.


          Let' consider rotation about x-axis.


The volume is the area of the ring in vertical section x= const, integrated over (or along) x-coordinate
from  x= a  to  x= 2a.  


The area of the ring is   = .


Thus the volume is the integral of the function    from  "a"  to  "2a".


The antiderivative is  the function  F(x) = .


So, the final integral value is the difference  F(2a) - F(a), which is

     =  =  = .


Thus the volume of the body of revolution in this case is  ,  in terms of "a", assuming "a" is positive.    ANSWER

Problem 2

The parabola  y = x^2  and straight line  y = x+2  have two common points (x,y) = (-1,1) and  (x,y) = (2,4).

You can determine it visually by plotting equations in your graphing calculator, or by solving equation x^2 = x+2.

Also, in the interval -1 < x < 2, the line is everywhere above the parabola, 
and both the line and the parabola are above x-axis.


You may check/prove it from inequality x+2 > x^2.


In any case, the area bounded by these curves, is between  y= x^2 from the bottom  and  y = x+2  at the top, 
in the interval  -1 <= x <= 2.


So, this area is rotated about x-axis, and they want you find the volume of the solid body formed.


Apply the disk method.

You may consider this volume as collection of the set of disks, or, better to say, as collection of rings.

Each ring is a vertical section of the body of rotation at x = const for some value of x between -1 and 2.

Each such disk has the outer radius of  x+2  and the internal radius of  x^2, so the area of each such ring is 


     = .


To find the volume, we should integrate this expression for the area of a ring from -1 to 2.


The antiderivative is  F(x) = .


Therefore, the integral is  F(2) - F(-1).


We have F(2)  =  = ,

        F(-1) =  = .


Thus the integral is  F(2) - F(-1) =  = .


ANSWER.  The volume of the solid formed is   =  = 45.2389  (rounded).

Problem 3

The outer boundary is y = 8-x^2.

The inner boundary is y = x^2.


These two curves intersect when  

    8-x^2 = x^2,  8 = x^2+x^2,  8 = 2x^2,  x^2 = 8/2 = 4,  x =  = +/- 2.


In the interval  -2 <= x <= 2, both curve are above the x-axis.


Each vertical section  x= const  of this volume  (of this solid body)  is a RING 
with the inner radius  r= x^2  and the outer radius R = 8-x^2.


The area of this ring is   =  = .


The antiderivative is  of this function    is  F(x) = .  


The integral from  -2  to  2  is the difference  

    F(2) - F(-2) = () - () =  =  =  = .


Thus the desired volume is   =  = 536.1651  (rounded).    ANSWER