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 description. 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 Check out our web site! http://mathforum.org/dr.math/
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