Applications of Integration to Geometry


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1 Applications of Integration to Geometry Volumes of Revolution We can create a solid having circular crosssections by revolving regions in the plane along a line, giving a solid of revolution. Note that the volume of a crosssection of such a solid is V πf(x) 2 x We can use integration to calculate the total volume of such a solid. Example 1 Calculate the volume of the solid created when the region bounded by the curve y e x, the xaxis, and the lines x and x 1 is revolved around the xaxis. Solution. π(e x ) 2 dx πe 2x dx pi 2 e x 1 pi ( ) 1 e 2 2 Sometimes our regions will not be completely solidthey may have holes cut out in the middle of them. Note that the crosssections of such regions will be like CD s. The area of a CD with radius, R, whose hole in the center has radius r is A πr 2 πr 2 Which can be simplified as A π(r 2 r 2 ) 1
2 Knowing the area, we can say that the volume of a crosssection of the solid with width x is V π(r 2 r 2 ) x Example 2 Calculate the volume of the solid created when the region bounded by the curves y x and y x 2 is rotated about the line y 3. Solution. Drawing the solid of revolution reveals that there is a hole through the middle of the object. Therefore, we know that we should use the formula V b a π(r2 r 2 )dx. We have a lot of things to identify here: 1) What is a? 2) What is b? 3) What is R? 4) What is r? We can find a and b by setting our two functions equal, finding the endpoints of the region they bound: x 2 x x 2 x x(x 1) x, x 1 That is, a and b 1. We must now find R and r. In general, R is the distance between the axis of revolution and the function which is further away from it. Now, r is the distance between the axis of revolution and the function which is closer to it. That is, R 3 x 2 and r 3 x Our volume is then V π[(3 x 2 ) 2 (3 x) 2 ]dx 2
3 π[(9 6x 2 + x 4 ) (9 6x + x 2 )]dx π[x 4 7x 2 + 6x]dx π[ x5 5 7x x2 ] 1 π[ ] 2.72 Sometimes we might revolve around a vertical axis instead of a horizontal axis but the same rules applyjust use horizontal rectangles instead of vertical rectangles. Example 3 Find the volume of the solid created when the region bounded by the curves y x 2 and y x is revolved around the line y 1. Solution. Since we are revolving around a vertical axis, we want to slice the region into horizontal rectangles. In order to do this, we need the functions in terms of y rather than in terms of x. So, the region we are revolving is bounded by the curves x y and Again, we must find a, b, R, and r. x y y y y, y 1 Thus, a and b 1. Now we find R and r: R y ( 1) and r y ( 1) 3
4 Then we can calculate the volume: V π[( y + 1) 2 (y + 1) 2 ]dy π[(1 + 2 y + y) (1 + 2y + y 2 )]dy π[ y 2 y + 2 y]dy π[ y3 3 y y3/2 3 ] 1 π[ ] π 2 Note that there is a way to do the previous problem without switching the direction of the rectangles. If you still wish to use vertical rectangles, then revolving them around the line y 1 gives you a very thin cylinder (think of a CocaCola can with no top or bottom. The volume of such a Coke can having width x is given by V 2πr(x)h(x) x Where r(x) is the radius of the can, and h(x) is the height of the can, given by the top function minus the bottom function. This method of using cylindrical shells is appropriately called the Shell Method: V b a 2πr(x)h(x)dx Example 4 Calculate the volume of the solid from the previous example, by using the shell method. Solution. We will use the shell method here, so we need to know a, b, r(x), and h(x). We find a and b as usual, by finding the intersection of our bounding functions: x x 2 4
5 x, x 1 We also need r(x), which can be found by taking the distance between the axis of revolution and the location of our shell: r(x) 1 x And now, h(x) is the height of a shell: Now the volume, V h(x) x x 2 2π[(x ( 1))(x x 2 )]dx 2π x x 3 dx 2π[ x2 2 x4 4 2π[ ] 1 π 2 Volume of Solids with Known CrossSections We don t always need to revolve shapes around axes in order to obtain volume. We might know what a crosssection of that solid looks like. Consider the following: Example 5 Calculate the volume of the solid whose base in the xyplane is bounded by the curves y x 2 and y 8 x 2, and whose crosssections are squares perpendicular to the xaxis. 5
6 Solution. We know that the crosssections of our solid are squares. What is the length of a side of one of these squares? S(x) (8 x 2 ) (x 2 ) 8 2x 2 The volume of a slice of width x of this solid will be V (8 2x 2 ) 2 x. Setting the two functions equal helps us find the limits of integration: x 2 8 x 2 x 1, x 2 Now, using what we know about integrals and volumes, we can find the volume of such a shape: V (8 2x 2 ) 2 dx 64 32x 2 + 4x 4 dx 64x 32x3 + 4x Arc Length We ve looked at calculating areas and volumes by chopping regions and solids into small pieces. We can do the same with calculating the length of a curve. Recall that the distance between two points, (x 1, y 1 ) and (x 2, y 2 ), in the plane is given by d (x 2 x 1 ) 2 + (y 2 y 1 ) 2. We can rewrite this as d ( x) 2 + ( y) 2 ( x) 2 + (f (x) x) 2 (1 + (f (x)) 2 )( x) 2 6
7 1 + (f (x)) 2 x Letting x, and adding together the distances over the interval [a, b] gives us the arclength formula: L b a 1 + (f (x)) 2 dx Example 6 Find the length of the portion of the curve y x 3 over [, 5]. Solution. Since f(x) x 3, then f (x) 3x 2 : L (3x2 ) 2 dx The integrals generated in trying to find arc length often require numerical methods to integrate. We can also calculate the arclength of a curve which is defined parametrically. For a particle whose motion is described by the parametric equations: { x(t) y(t) Then the distance the particle travels over the time interval [a, b] (also known as the arc length) is given by b a (x (t)) 2 + (y (t)) 2 dt Example 7 Set up the integral to find the circumference of the ellipse given by the equations { x(t) 2 cos(t) for t 2π. y(t) sin(t) 7
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