LESSON 14: VOLUME OF SOLIDS OF REVOLUTION SEPTEMBER 27, 2017

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1 LESSON 4: VOLUME OF SOLIDS OF REVOLUTION SEPTEMBER 27, 27 We continue to expand our understanding of solids of revolution. The key takeaway from today s lesson is that finding the volume of a solid of revolution is all about determining the radius. Specifically, we will be revolving around vertical and horizontal lines that are not the y- and x-axes where we need to think very geometrically about how we determine the radii of our disks. Examples.. Consider the region bounded by the curves y =, y =, x =, x =. x a) Find the volume of the solid generated by revolving the region about the line y =. We sketch a graph to get a geometric understanding of what is going on. We are not revolving around the x-axis, but instead the line y =. So, we need to be more careful with how we choose our outer radius and inner radius because now it is relative to y = and not the x-axis.

2 2 MATH 62 Our outer radius is the difference between y = and y = and the inner radius is the difference between y = and y = think about x how you would draw the disks and determine their radii). Then we use exactly the same formula as we have been using for the washer method: Volume = π ) 2 ) ] 2 dx. x So, Volume = = π = π π 2 2 x ] dx x 2 ln x + x ] = π ln + x + )] dx x 2 ) = π ln ln ) + 2] = π ln 2 ] ln + )] b) Find the volume of the solid generated by revolving the region about the line x =. We go about this with very much the same spirit as in part a). However, there are slight differences. Consider the graph

3 MATH 62 3 We observe that depending on the y-value, we have different outer radii. So we need to break the graph into Area and Area 2. We do want to observe that for either Area or Area 2, the inner radius is because there s no gap between where we are revolving and what we are revolving. So we use the disk method for both areas. Since we are revolving around the a vertical line, we need to solve for x. Given y =, we get x =. Thus, for Area, x y

4 4 MATH 62 we get For Area 2, Radius : x = y Bounds : y 2 we get Radius : x = Bounds : y Remark. These radii are the differences between the functions and the line we are revolving about. Therefore, our volume is given by Volume = π ) 2 + }{{} = Area 2 π) 2 + = π 2 + = π 2 2 2y + y ) 2 π y }{{} Area ) 2 2 π y y 2 y + 2 ln y + 2y = π 2 ln ) 2 = π 2 ln ] = π ln 2] = π ln 2) ) ] ) ] 2 ln + 2) )]

5 MATH Let S be the region bounded above by x 4 y = 6, below by y =, on the left by x =, and on the right by x = 2. a) Find the volume of the solid generated by revolving S around the line y =. Because we are revolving about the line y =, we are revolving about a horizontal line, which means our radius function must be a function of x i.e., a function that that spits out y-values). Given x 4 y = 6, solving for y we get y = 6 x 4. Now, the radius is given by 6 and the bounds are x 2. x4 Note that we don t need to use the washer method here because there is not space between the region and the line we are revolving about. So our volume is given by 2 ) 2 6 Volume = π x dx 4 = π 2 26 x 8 32 x 4 + ) dx

6 6 MATH 62 = π 26 = π ] 2 3x + x 3 7x ) ) = 87π 2 ) 26 7) + 32 )] 7 3) + 3 b) Find the volume of the solid generated by revolving S around the line x = 2. Here, we are revolving around a vertical line. This means that our radius must be a function of y i.e., a function that spits out x-values). So we solve for x: x 4 y = 6 x 4 = 6 y Thus, our graph looks like x = 4 6 y x = 2 y /4 This does require the washer method because there is a gap between the region and the line we are revolving about.

7 MATH 62 7 This means Outer Radius : x = 2 Inner Radius : x = 2 2 y /4 Bounds : y 6 Therefore, our volume is given by 6 Volume = π 2 ) ) ] 2 y /4 6 = π ) ) ] 2 y /4 6 = π y 8 )] /2 y /4 6 = π 3 4 y + 8 ) /2 y /4 = π 3y 8y / ] 6 3 y3/4 = π 36) 86) / )3/4 = π 48 84) ) = 7π 3 3) 8) / Find the volume of the solid generated by the region enclosed by about the line x =. y = x, y =, x =, x = As usual, we sketch a graph of this region. ] 3 )3/4 )]

8 8 MATH 62 This is certainly a problem involving the washer method since there is a gap between the region and where we are revolving. Moreover, the inner radius differs depending on what y-value we choose. Thus, we need to divide this region into two areas: We also want to note that because we are revolving about a vertical line, we need to have radius functions which provide x-values. So, we solve for x: y = x x = y. For Area, we see that

9 MATH 62 9 Outer Radius : x = Inner Radius : x = y) Bounds : y and for Area 2, we see Outer Radius : x = Inner Radius : x = Bounds : y We put this together to compute the volume: Volume = π ) 2 ) 2] + π ) 2 y)) 2] }{{}}{{} = Area 2 π ) 2 ) 2] + Area 2 π ) 2 + y) 2]

10 MATH 62 = π 22 ) + = π 2) y + y 2 ) ) ] ] 2 y y 2 ) = π 2y + 2y 2 y2 ] 3 y3 = π 2) + 2) ) 2 3 )3 2) ) 2 3 ))] )3 = π 62 + ] = 287π 3 3