Volume of a Rotated Region
Date: 11/02/98 at 00:19:38 From: Marianne Subject: Calculus Find the volume of a solid formed by revolving the region bounded by y = x^2 + 1, y = 0, x = 0, and x = 1, about the y-axis.
Date: 11/02/98 at 09:11:18 From: Doctor Jaffee Subject: Re: Calculus Hi Marianne, The region you are describing is below the parabola in the first quadrant. Draw a picture of the parabola and the vertical lines x = 0 and x = 1. Within that region draw a thin rectangle whose base is on the x-axis and whose opposite base intersects the parabola. If you were to revolve this rectangle around the y-axis it would sweep out a region similar to that of a tin can with no top or bottom. The volume of metal could be determined by slicing the can vertically and opening it up into the shape of a rectangle. So, the volume of the rectangular solid of tin would be length times width times the thickness of the sheet of metal. But the length of the rectangle is just the circumference of the circular top of the tin can, the width of the rectangle is the height of the can, and the thickness is a very tiny quantity that we can call "dx". Furthermore, the circumference of the circle is just pi * r^2, where r is the x number at the base of the original rectangle. Also, the height is the distance from the x-axis to the y number on the parabola, (x^2 + 1). So, the volume of the tin on our tin can is pi * x^2 * (x^2 + 1) * dx. Now, if we were to place infinitely many, infinitely thin rectangles in our region and revolve them all around we would have the entire solid of revolution. The volume of this solid would be the sum of each of the individual revolutions of these thin rectangles, or: integral evaluated from 0 to 1 of pi * x^2 * (x^2 + 1)dx I'll leave it up to you to do the integration. I hope this explanation has helped. It was very difficult to explain without using diagrams, but do try to follow my text, draw your own picture, perhaps consult with some friends and teachers, and figure out what I'm trying to say. Good luck. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/
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