We imagine the egg being the three dimensional solid defined by rotating this ellipse around the x-axis:

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1 CHAPTER 6. INTEGRAL APPLICATIONS 7 Example. Imagine we want to find the volume of hard boiled egg. We could put the egg in a measuring cup and measure how much water it displaces. But we suppose we want to do it more mathematically, using functions and formulas. We will imagine cutting the egg into slices, and then measuring the volume of each slice as a cylinder. (a) Use the ellipse x + y.8 = to model the outline of an egg with length.3 in and diameter.6 in. Draw the ellipse and the shape it makes when rotated around the x-axis. (b) Imagine cutting the shape from (a) into slices, and draw the result. (c) Imagine replacing each slice from (b) with a cylinder and draw the result. (d) Figure out how to find the volume of each cylinder-slice in (c) and write down a formula for the sum of the volumes. (e) Translate the formula from (d) into an integral, and solve the integral. Solution. (a) We imagine the egg being the three dimensional solid defined by rotating this ellipse around the x-axis: (b) Show how we can slice this volume into pieces, and approximate the volume of each piece using a disk, and then turn the the sum of volumes of disks into an integral. We start by drawing a picture of the egg after it s cut into a bunch of slices (notice that cutting a shape into pieces doesn t change the total volume): Vol

2 CHAPTER 6. INTEGRAL APPLICATIONS 8 (c) Now each slice has a slight curve on the top and bottom. This makes it hard to find the exact volume of the slice, but we can approximate the volume by using a cylinder of the same size: Vol (d) Now, we can give a formula for the volume of all these disk-shaped slices above stuff r x + πr x + πr 3 x +... To find r, r,etc.,wegobacktohowthisshapeisdefined:theradiusisthe distance of the curve from the x-axis. This distance is what we usually call the x y-value, i.e. r = y. Now we find the y-values from the ellipse + y.8 = by solving for y: y x =.8 y =.8 x y =.8 x y =.8 x Combining this formula with the above sum of volumes we get above stuff.8 x / x+π.8 x / x+π (e) Now we apply the Approximation Principle: anything that we can approximate with a sum of function values, we can make exact by integrating: π.8 x / dx This last line is the most important conclusion of this example: translating avolumebyrotationintoanintegral. Ofcourse,it salsonicetofindthe integral: π.8 x / dx.8 x / dx (.8) x / dx.8 x 3/ x+...

3 CHAPTER 6. INTEGRAL APPLICATIONS 9 (.8) x =3.in 3 x () Example. Find the volume generated by rotating y = x + around the x-axis, between x = and x =. Solution. The original shape, and the volume generated by it are pictured below. r = y = x + x We apply b a π(f(x)) dx with a =, b =, f(x) = x +andget π( x +) dx x 4 4x +4dx 5 x5 4 3 x3 +4x 5 5/ 4 3 3/ / / 4 =π 5 5/ 4 3 3/ +4 =π = 4 π =8 3 π = 64 π 5 Example 3. Find the volume generated by rotating y = x around the y-axis, between y = and y =.

4 CHAPTER 6. INTEGRAL APPLICATIONS Solution. This is a sideways problem because the way we can cut it into disks is to make cuts that go up and down along the y-axis. One slice of this volume gives a disk of thickness y and and radius given by the x-value. To get the x-value we need to write x(y), i.e. x as a function of y. We solve y = x for x = y.wealsoneedtoknowtheboundsofintegration; this is from the smallest y-value to the largest one. The y-values are y = and y =. π( y) dy y =π ydy Example 4. Set up an integral, and use your calculator to find it, for the volume of the napkin ring made by rotating the region bounded by y = x +, and y = around the x-axis. Solution. volume: We picture the region defined below, along with one slice of

5 CHAPTER 6. INTEGRAL APPLICATIONS The volume of the slice is given by the area of the face, times x. The main work is to figure out the formula for the area. We start with the two dimensional shape, involving r and R r R Area R πr Volume = (πr πr ) x So, to finish, we just need to figure out formulas for R and r in the shape that we have. R r Thus, R = y = x +andr =. Puttingallofthistogether,wehave the integral that we want. π x + π dx = π x + π dx π( x +4 x +4 4) dx x +4 x +dx x dx

6 CHAPTER 6. INTEGRAL APPLICATIONS To finish this, we need to figure out what x dx is. This is probably not something that you know how to find the anti-derivative of. But, you can figure out the integral without knowing the anti-derivative. If you think of what area is represented by x dx, youshouldbeabletoseethat it s the area of the top half of a unit circle. This has area π.thus, 4 π 3 +4 π = 4 3 π +π Example 5. Find the volume of the napkin ring made by rotating the region bounded by y = 4 9 x +, and y = 4 x + 7 (shown below) around the line 6 y =. Solution. The rotated shape looks roughly as follows:

7 CHAPTER 6. INTEGRAL APPLICATIONS 3 graphics/rotated_graph_for_napkin_ring_of_quadratics-eps-converted-to.pdf The larger radius is defined by the top curve, y = 4 9 x +. It is the distance between this curve and the horizontal axis, y =. Thus, R = 4 9 x + ( ) = 4 9 x +3. The smaller radius, the radius of the hole, is defined by the bottom curve y = 4 x Itisthedistancebetweenthiscurveandthehorizontalaxis y =. Thus r = 4 x ( ) = 4 x Now we integrate (skipping some of the messy steps) 3/ π x π 4 x + 3 dx 6 3/ (messy steps of foiling skipped) = 5π 6 = 5π 6 3/ 3/ 3/ x x dx 3/ = 5π 6 87 = π 7 8 x4 3 6 x dx 7 45 x5 3 8 x x 3/ 3/

8 CHAPTER 6. INTEGRAL APPLICATIONS 4 Example 6. Find the volume of the region between the curves y = g(x) =x 3 and y = f(x) = x rotated around the line x =(shownbelow). Solution. We solve this with washers, that are stacked up and down along the vertical line x =. Thus,wewillhave b a πr πr dy where we need R and r to be formulas for the radiuses, written as functions of y. The larger radius, R, isthehorizontaldistancebetweenx =andthe curve f(x). We need the formula in terms of y, sowesolvey = x for x = y, R = y The smaller radius, r, is the horizontal distance between x =andthecurve g(x). Again we rewrite the equation first y = x 3 x = y /3 r = y /3

9 CHAPTER 6. INTEGRAL APPLICATIONS 5 Putting it all together we get π( y ) π( y /3 ) dy y + y 4 ( y /3 + y /3 ) dy y + y 4 +y /3 y /3 dy 3 y3 + y5 = 3 3 π +y4/3 5 4/3 y5/3 5/ Extra examples Example 7. Derive the formula for the volume of a sphere of radius r. Solution. We start with the formula for the top half of a circle of radius r: y = r x and then plug this into our volume by rotation formula r r r π( r x ) dx (r x ) dx r r x x3 r 3 r r r r3 3 =π r 3 r3 3 =π 3 r3 = 4 3 πr3 r r + r3 3 Challenge. Can you figure out how to find the volume of a shape rotated around the line y = x? What about other lines? Can you figure out how to apply washers when f(x) and g(x) switch places with respect to which one is farther from c? What about if they switch places also with respect to which side of c they fall on?