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Volume of a Shape

Date: 11/11/96 at 12:49:46
From: Daniel H. Friendly
Subject: Finding volume of a shape

Dear Dr. Math -

This may be beyond the scope of k-12, but perhaps the AP calculus 
courses cover it. In any event, I am stumped and would love it if you 
could help.  

What is the volume of the shape formed by rotating the parabola,
y=x^2 around the line y=x? (From x = 0 to 1). 

Date: 11/12/96 at 14:27:20
From: Dr. Donald
Subject: Re: Finding volume of a shape

This is a pretty nasty problem.  At least, I would never assign it.  
But you can figure it out using volumes by shells.

The basic formula here is integral (2 pi r h) dt where t ranges from 
one end of the solid to the other.  r is the distance of the boundary 
of the region to the axis of rotation and h is the height of the 
cylindrical shell.  dt is the thickness of the shell.  

A line parallel to y=x with y-intercept a has equation y = a + x 
and meets the parabola at the solutions to a+x = x^2, or 
x = (1 +- sqrt(1+4a))/2, where a ranges through negative values 
from 0 down to the point where the line y = a+x is tangent to the 
parabola.  This happens when the equation has only one solution, 
i.e. a = -1/4.

For a shell corresponding to the line y = a+x, the height is the 
distance between intersections with the parabola. The line segment 
between intersections is the hypotenuse of an isosceles right triangle 
whose leg equals the difference in the x-values of the intersections, 
i.e. the roots found above.  The difference of the roots is just
sqrt(1+4a), and so the hypotenuse is sqrt(2)sqrt(1+4a).

The radius of the shell at this line is the leg of another such 
isosceles right triangle, with hypotenuse a; this little triangle has 
one vertex at the origin, one at the point (0,a), and the other at the 
point (-a/2,a/2).  Remember that a is negative.  This triangle has leg 
of length -a/sqrt(2).  

Finally, if the shell lies between two such lines whose intercepts 
differ by da, then the lines are da/sqrt(2) apart for the same reason 
that we divided by sqrt(2) in calculating the radius.

So we now have the ingredients for the integral:

integral(2 pi (-a/sqrt(2)) sqrt(2)sqrt(1+4a) (1/sqrt(2))da
from -1/4 to 0).

You can collect all the factors and end up with

sqrt(2) pi integral(-a sqrt(1+4a) da from -1/4 to 0) 

You should make a careful drawing and make it agree with this 

It is possible to define more generally the volume swept out when 
rotating about an arbitrary line.  It is somewhat uglier than what we 
have done here but the same basic idea.  This is rarely discussed in 
calculus texts.

-Doctor Donald,  The Math Forum
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Associated Topics:
High School Calculus

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