4 More Applications of Definite Integrals: Volumes, arclength and other matters

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1 4 More Applications of Definite Integrals: Volumes, arclength and other matters Volumes of surfaces of revolution 4. Find the volume of a cone whose height h is equal to its base radius r, by using the disc method. We will place the cone on its side, as shown in the Figure, and let x represent position along its axis. (a) Using the diagram shown below (Figure ), explain what kind of a curve in the xy plane we would use to generate the surface of the cone as a surface of revolution. y generating curve r x h Figure : For problem 4. (b) Using the proportions given in the problem, specify the exact function y = f(x) that we need to describe this curve. (c) Now find the volume enclosed by this surface of revolution for x. (d) Show that, in this particular case, we would have gotten the same geometric object, and also the same enclosed volume, if we had rotated the curve about the y axis. (e) Suppose you are now given a new line, y = ax and told to compute the two volumes as in (c) and (d) above. By what factor would they differ? (a) (b) The curve that generates the cone surface of revolution is the straight line y = r h x = x (since r = h). (c) Disk method - rotate about the x axis. Each disk has width dx and radius y. Volume of each disk is (area of base) (width) = πy dx. Because the curve used to generate this surface of revolution is y = f(x) = x, the volume of a disk is π[f(x)] dx = πx dx.

2 Summing up the volumes enclosed by each of the disks, we find the total volume enclosed by this surface of revolution for < x < : V = π[f(x)] dx = πx dx = π/3 cubic units (d) Rotate about the y axis: since r = h, we obtain the same geometric object. In this case, each disk has width dy and radius x. The volume of each disk is therefore π[f(y)] dy = πy dy, since f(y) = y. Volume enclosed by surface of revolution is then for < y < and is (same as in part (c)). π[f(y)] dy = πy dy = π/3 cubic units (e) The two volumes are: a π/3 and a π/3 respectively. Thus, they differ by a factor a Find the volume of the cone generated by revolving the curve y = f(x) = x (for < x < ) about the y axis. Use the disk method, with disks stacked up along the y axis. Figure : The disks have thickness y dy and radii in the x direction so that x = f(y) = y is the function to use. From the interval < x <, we find the interval on the y-axis is < y <. Thus, V = π [f(y)] dy = π [ y] dy

3 4.3 ( ) V = π [ y + y ] dy = π y y + y3 3 V = π( + 3 ) = π 3. Find the volume of the bowl obtained by rotating the curve y = 4x about the y axis for x. We are rotating about the y axis, so, using the disk method, disk width is dy, the radius of each disk is in the x direction, i.e. it is given by x = f(y) = y/4. The volume will be given by V = ymax y min π[f(y)] dy. To find y max and y min we use the interval x. But y() = and y() = 4. Therefore the interval on the y-axis is y π[f(y)] dy = π 4 4 y dy = π. On his wedding day, Kepler wanted to calculate the amount of wine contained inside a wine barrel whose shape is shown below in Figure 3. Use the disk method to compute this volume. You may assume that the function that generates the shape of the barrel (as a surface of revolution) is y = f(x) = R px, for < x < where R is the radius of the widest part of the barrel. (R and p are both positive constants.) y y = R px x Figure 3: For problem 4.4 3

4 radius r=r px^ dx Figure 4: For problem 4.4 solution x, R >, p >. Disk method: See Figure 4. The radius of each disk is y = R px, and the thickness is x dx. The volume of each disk is therefore V = π(r px ) x Because f(x) = R px is an even function, the volume for < x < is twice that for < x <. Volume = π (R px ) dx = π (R Rpx + p x 4 ) dx ( ) = π R x Rp x3 x5 + p 3 5 ( = π R Rp ) 3 + p 5 So the amount of wine in the barrel is π ( R RP 3 ) + p cubic units Consider the curve y = f(x) = x < x < rotated about the y axis. Recall that this will form a shape called a paraboloid. Use the cylindrical shell method to calculate the volume of this shape. The cylindrical shells will encase one another with their radii in the direction of the x axis, their heights along the y axis, and their thickness x dx along the x axis. The radius of 4

5 a given shell is x, the height is y = f(x) = x. The volume of one shell is V shell = πrhτ = πxf(x) dx The volume of the entire shape for < x < is 4.6 V = π xf(x) dx = π x( x ) dx. ( ) x V = π (x x 3 ) dx = π x4 ( = π 4 4) Consider the curve y = f(x) = x (which is a part of a circle of radius ) over the interval < x <. Suppose this curve is rotated about the y axis to generate the top half of a sphere. Set up an integral which computes the volume of this hemisphere using the shell method, and compute the volume. (You will need to make a substitution to simplify the integral. To do this question, some techniques of anti-differentiation are required.) The generating curve is y = f(x) = x. We can consider a typical shell of thickness x whose height is determined by the value of y on the curve. Then the radius of the shell is r = x, the height is y = f(x) = x, and the volume is V shell = πx x x. The integral (sum of all shell volumes as x ) is V = π x x dx. To compute the integral, let u = x. Then du = dx so = π V = π u / du = π u3/ 3/ V = π 3 ( x ) 3/ = 3 π. Since the radius of the generating circle is and we have half of a sphere, the volume is (/)(4/3)πr = (/3)π. 5

6 4.7 Find the volume of the solid obtained by rotating the region bounded by the given curves f(x) and g(x) about the specified line. (a) f(x) = x, g(x) =, from x = to x = 5, about the x-axis. (b) f(x) = x, g(x) = x/, about the y-axis. (c) f(x) = /x, g(x) = x 3, from x = / to x =, about the x-axis. (a) The volume of the solid is equal to 5 π(x )dx = π(x / x) 5 = 7.5π. (b) To calculate the volume of the solid, we have to find the x-coordinates of the curves as a function of the y-coordinates, and we have to determine the boundaries on the y-axis. For the first task, we simply take the inverse functions: x = y and x = y. Thus, the solid is obtained by rotating the region between the curve x = y and the curve x = y around the y-axis (see Figure 5). This region is bounded by the two points (x, y) = (, ) and (x, y) = (4, ) (the latter point can be found by solving f(x) = g(x)). Therefore, we have to integrate from to on the y-axis. From a sketch, we see that the volume is a paraboloid V with a conical hollow core V. The volume can be computed by subtraction V = V V. We get the volume of the solid as V = V = π(y) dy π(y ) dy π[(y) (y ) ]dy V = π( 4y3 3 y5 5 ) = 64π 5 (c) Since f(x) g(x) in the region considered, the volume of the solid is V = (πf(x) πg(x) )dx = π V = π( /x x 7 /7) (x x 6 )dx = π(6 + / 7 )/ π. 6

7 F(x) X 4.8 Figure 5: For problem 4.7 solution Let R= the region contained between y = sin(x) and y = cos(x) for x π/. Write down the expression for the volume obtained by rotating R about (a) x-axis (b)the line y=-. (a) V = π 4 π(cos (x) sin (x)) dx + π π π(sin (x) cos (x)) dx = π 4 π(cos (x) sin (x)) dx 4 = π 4 π cos(x) dx (b)v = π 4 π[( + cos(x)) ( + sin(x)) ] dx + π π π[( + sin(x)) ( + cos(x)) 4 π 4 π[( + cos(x)) ( + sin(x)) ] dx 4.9 Find the volume of a solid torus (donut shaped region) with radii r and R as shown in Figure 6. (Hint: There are several ways to do this. You can consider this as a surface of revolution and slice it up into little disks with holes ( washers ) as shown.) torus ] dx R r "washer" (disk with a circular hole) Figure 6: For problem 4.9 7

8 y line x=r x R Figure 7: For problem 4.9 solution dx y=(r^ (x R)^)^(/) rotating about the y axis cylindrical shell x dx Figure 8: For problem 4.9 solution: shell Consider the upper semi-circle (shaded region in Figure 7) rotated about the y axis. The volume of the torus is then times the resulting volume of revolution. The circle centered at (R, ) with radius r is (x R) + y = r, so the upper semi-circle is y = r (x R) for R r x R + r. One Way: See Figure 8. Volume of one shell = (width)(height)(circumference) = dx r (x R) πx. R+r Volume of torus = R r πx R+r = 4π xr R r r (x R) dx ( ) x R dx r Let sin u = x R x = R + r sin u, dx = r cos u du. Notice that the limits of integration r now change to π/ u π/. π/ Volume = 4π [r(r + r sin u) cos u r cos u] du π/ 8

9 x = R (r^ y^)^(/) x = R+ (r^ y^)^(/) "washer" dy rotating about the y axis for <y<r dy = 4πr π/ = 4πr R = 4πr R = πr R Figure 9: For problem 4.9 solution: washer π/ π/ π/ π/ π/ ( u + ( ( π = πr R π = π r R cubic units cos u(r + r sin u) du cos u du + 4πr 3 π/ + cos u du + ) sin u π/ u= π/ )) π/ cos u sin u du Another Way: See Figure 9. For this method we use the disk method, rotating about the y-axis. Volume of one washer = (volume of solid disk) (volume of hole) = πrdiskh πrholeh. The radius of the solid disk is equal to the equation for the right-hand side of the circle in Figure 7. This equation, as shown in Figure 9, is r disk = R + r y. Similarly, the radius of the hole is the defined by the equation that describes the innermost part of the circle, the left-hand side as seen in Figure 7 and Figure 9. Thus r hole = R r y. Combining this with the height of each washer (dy), we get an equation for the volume of one washer: V washer = π(r + r y ) dy π(r r y ) dy. To find the volume of the torus, we would integrate this over r y r. But because the function is symmetric about the x-axis, we instead integrate over y r, and double the result. Volume of torus = = π r r [ π (R + ] r y ) (R r y ) 4R r y dy 9 dy

10 r = 8πR r y dy ( ) πr = 8πR (which is 4 4 = π Rr cubic units of the area of a circle of radius r) 4. In this problem you are asked to find the volume of a height h pyramid with a square base of width w. (This is related to the Cheops pyramid problem, but we will use calculus.) Let the variable z stand for distance down the axis of the pyramid with z = at the top, and consider slicing the pyramid along this axis (into horizontal slices). This will produce square slices (having area A(z) and some width z). Calculate the volume of the pyramid as an integral by figuring out how A(z) depends on z and integrating this function. z h w Figure : For problem 4. Let the side length of the square cross-section at height z be x. Then by similar triangles, x/z = w/h, or x = (w/h)z. Now the area of the square slice at height z is A(z) = x = (w/h) z and its thickness is z. Thus V = h (w/h) z dz = (w/h) [z 3 /3] h = w h/3.

11 Length of a curve and arclength 4. Set up the integral that represents the length of the following curves: Do not attempt to calculate the integral in any of these cases (a) (b) (c) y = f(x) = sin(x) < x < π. y = f(x) = x < x <. y = f(x) = x n < x <. (a) dy/dx = cos(x) so (b) y = f(x) = sin(x) < x < π. L = π + cos (x) dx y = f(x) = x < x <. dy/dx = x /, and (dy/dx) = ( ) x = 4x so L = + 4x dx (c) dy/dx = nx n so y = f(x) = x n < x <. L = + n x n dx

12 4. Compute the length of the line y = x + for < x < using the arc-length formula. Check your work by using the simple distance formula (or Pythagorean theorem). dy/dx = so L = L = 5 + (dy/dx) dx = dx = 5x + 4 dx = 5( ( )) = 5. We can compare this to the simple distance formula: The endpoints of the line are (, ) and (, 3) so the length of the line is d = ( ( )) + (3 ( )) = + 4 = = Find the length of the curve y = f(x) = x 3/ < x < (To do this question, some techniques of anti-differentiation are required.) We use the arclength integral for the curve y = f(x) = x 3/. To do so, we first find Then the length we want is and, simplifying, L = f (x) = 3 x/. + (f (x)) dx = + ( 3 x/ ) dx, L = x dx. Substitute u = + (9/4)x. Then du = (9/4)dx, or dx = (4/9)du. Plugging into the integral leads to L = 4 x= 4 u 3/ u du = 9 x= 9 3/ = 4 ( + (9/4)x) 3/. 9 3/ We compute L = 8 7 ( ( + (9/4)) 3/ ) =

13 4.4 To do this question, some techniques of anti-differentiation are required. (a) Use the chain rule to show that F (x) = sec(x) is the derivative of the function F (x) = ln(sec(x) + tan(x)) So, what is the anti-derivative of f(x) = sec(x)? (b) Use integration by parts to show that sec 3 (x)dx = (sec(x) tan(x) + ln(sec(x) + tan(x)) + C. You will want to recall the trigonometric identity + tan (x) = sec (x). (c) Set up an integral to compute the arclength of the curve y = x / for < x <. (d) Use the substitution x = tan(u) to reduce this arclength integral to an integral of the form sec 3 (u) du. (a) Differentiating leads to F (x) = sec(x) + tan(x) (sec(x) tan(x) + sec (x)) where we have used the chain rule and derivatives of sec(x) and tan(x). Now cancelling the common factor leads to F (x) = sec(x). From these results, we see that sec(x) dx = ln(sec(x) + tan(x)). (b) Let u = sec(x) du = sec(x) tan(x)dx. Let dv = sec (x)dx v = tan(x). Then I = sec 3 (x)dx = sec(x) tan(x) sec(x) tan (x) dx Using tan (x) = sec (x) we get I = sec(x) tan(x) sec(x)(sec (x) ) dx 3

14 Multiplying through gives sec 3 (x)dx = sec(x) tan(x) (sec 3 (x) sec(x)) dx. sec 3 (x)dx = sec(x) tan(x) sec 3 (x) dx + sec(x) dx. sec 3 (x)dx = sec(x) tan(x) + ln(sec(x) + tan(x)) sec 3 (x) dx. We can move the sec 3 (x) dx to the other side and divide by to get sec 3 (x)dx = (sec(x) tan(x) + ln(sec(x) + tan(x)) (c) y = f(x) = x / so f (x) = x. The length of the curve is then L = + x dx. (d) Let x = tan(u). Then dx = sec (u)du. We also use the fact that + tan (u) = sec (u) so that the integral becomes 4.5 L = + x dx = sec (u) sec (u) du = sec 3 (u)du. Second Spreadsheet assignment: You are told that the derivative of a certain function is f (x) = sin (x/3) and that f() =. Use the spreadsheet to create one plot that contains all of the following graphs: () The graph of the function y = f(x) (whose derivative is given to you). This should be plotted over the interval from to 9. () The graph of the element of arclength dl = + (f (x)) dx showing how this varies across the same interval. (3) The graph of the cumulative length of the curve L(x). Briefly indicate what you did to find the function in part (). (You might consider how the spreadsheet would help you calculate the values of the desired function, y = f(x), rather than trying to find an expression for it.) Here is how we could organize the spreadsheet: We tabulate x values (in increments of x =. in the a column, and values of the given f (x) in the b column. We then compute f(x) in the c column by the linear approximation formula f(x + x) = f(x) + f (x) x. 4

15 . Arc length L(x) - total length f(x) dl (element of arc-length) V -. the given f (x) (optional). 9. Figure : Thus, we start with f() = in cell c, and set cell c to c+b*., computing the value of f at this place, and so on. (Drag the entry down to generate the whole column.) In column d we calculate dl = + (f (x)) dx by setting cell d to sqrt( + b )., and dragging the entry down. We add up these values in column e by setting e =, e = e + d, and so on, to get the total length L(x) as it accumulates along the curve. The resulting plot is shown in Figure. We also show an expanded version (which makes dl clearer to see) in Figure. Other applications 4.6 work A spring has a natural length of 6 cm. When it is stretched x cm beyond that, Hooke s Law states that the spring pulls back with a restoring force F = kx dyne, where the constant k is called the spring constant, and represents the stiffness of the spring. For the given spring, 8 dyne of force are required to hold it stretched by cm. How much work (dyne-cm) is done in stretching this spring from its natural length to a length 4 cm? 5

16 .5 V.. 9. Figure : Since F = 8 when x = we find that 8 = k so that k = 4 and F = 4x is the force. When the spring is at its natural length, 6 cm, no work is done. To stretch the spring by x beyond this would require work W = F x = 4x x. To sum up all the work involved in stretching from 6 to 4 cm, i.e. for the increase in length x = to x = 8, we would calculate 4.7 work W = F dx = 8 4xdx = x 8 = 8 dynes cm. Calculate the work done in pumping water out of a parabolic container. Assume that the container is a surface of revolution generated by rotating the curve y = x about the y axis, that the height of the water in the container is units, that the density of water is gm/cm 3 and that the force due to gravity is F = mg where m is mass and g = 9.8m/s. The paraboloid can be thought of as a bowl full of water. The total work done is equivalent to the sum of work done to pump each bit of the water up to the height, or equivalently, 6

17 a sum over disks of water, each at some height y. The work done is Work = (force) (distance). The distance from a disk at height y to height is y. Therefore, Work = Force ( y). The force is F = mg where g = 9.8m/s = 98cm/s. The mass of a disk at height y is m y = ( gm/cm 3 )(πr y cm 3 ) where the radius of the disk is the x coordinate of a point on the parabola y = x, giving r = x = y. Thus, the work done to move this disk to height cm is w y = m y g( y) = π( y) y(98)( y) = 98π( y)y y, and the total work, which is a sum over all disks, is the integration of w y over < y < : W = 98π W = 98π ( y)y dy = 98π (y y ) dy ( ) y y3 3 (One erg is gm cm s or dyne-cm). 4.8 Energy ( = 98π 3 ) = ergs. 3 In a harmonic oscillator (such as a spring) the energy is partly kinetic (associated with motion), and partly potential (stored in the stretch of the spring). Energy conservation implies that the sum of the two forms (E) is constant so that mv + kx = E where m is the mass on the spring, k the spring constant. techniques of anti-differentiation are required.) (To do this question, some (a) Use the fact that the velocity is v = dx/dt to show that this energy conservation equation implies that ( dx dt = m E k ) / m x. (b) Rewrite this in integral form by dividing both sides of the equation by the term in the square-root to obtain dx ( E ) / = dt. k m m x (c) Integrate both sides. You will need to use substitution, and the result will be an inverse trig function for x. 7

18 (d) Find the displacement x as a function of time t. Show that the motion of the spring is periodic. (a) The velocity v = dx/dt so that ( ) dx m + ( ) dx dt kx = E, so that m = E kx. dt Simplifying gives dx ( ) E kx / dt =. m dx (b) ) / = dt is obtained by cross multiplying and integrating both sides, and ( E kx m is called separation of variables. (c) We simplify to Let a = E/k. Then the above is k m dx (E/k) x = t + C. m k trig-type integral, and we get m k sin (x/a) = t + C. dx = t + C. The integral on the left is a a x m k (d) k sin (x/a) = t + C, so sin (x/a) = (t + C), i.e (x/a) = sin k (t + C). m m Thus x = a sin k (t + C) E = m k sin k (t + C). This is a periodic function with m k E frequency ω = and amplitude m k. 4.9 Shear A vertical shear is a transformation of the plane which takes the vertical line x = a and pushes it upward by a distance sa (the constant s is called the strength of the shear). For instance, the rectangle shown below in Figure 3 is transformed into a trapezoid. (a) Set up a definite integral which shows that the area of the trapezoid is the same as the area of the original rectangle. We say that the shear has preserved the area of the rectangle. (b) Now consider a more general shape: Show that the area before the shear is the same as the area after the shear. (a) See Figure 5. 8

19 y Before shear y After shear x = a x x = a x Figure 3: For problem 4.9 Before Shear After shear Figure 4: For problem 4.9(b) Area of rectangle = ab. Area of trapezoid = a [(sx + b) sx] dx = a b dx = bx a = ab. So shear is area-preserving for a rectangle. (b) See Figure 6. Consider the circle as made up of rectangle strips, each of which is undergoing a shear of strength s. From part (a) we know that this shear is area-preserving. Therefore, the area of the circle (before shear) is the same as the area of the ellipse (after shear). 4. Surface Area Calculate the surface area of a cone shaped surface obtained by rotating the curve y = x on the interval [, ] around the x-axis. To do this question, some techniques of antidifferencation are required. Using the formula for surface area S = πy(x) + (y (x)) dx. 9

20 y b Before shear y=b y b After shear y = sx + b line x = a x = a x y = sx a sa x Figure 5: For problem 4.9 (a) solution a a a a a a a Circle (before shear) a Ellipse (after shear) Figure 6: For problem 4.9 (b) solution Now, y = x and y (x) = /( x). Thus S = π x + 4x dx = π π 4x + dx = (4x+) / d(4x+) = π 4 (4x+)/ = π [3 ] = π.