Chap 10 Applications of Definite Integral

Chap 10 Applications of Definite Integral - FIRST YEAR...

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Unformatted text preview: FIRST YEAR CALCULUS W W L CHEN c W W L Chen, 1994, 2005. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 10 APPLICATIONS OF THE DEFINITE INTEGRAL 10.1. Areas on the Plane Recall that in Chapter 7, the Riemann integral B f (x) dx A is formulated in terms of the area bounded by a curve y = f (x) and the lines y = 0, x = A and x = B . In this section, we shall use the same idea to help us evaluate areas on the plane. First of all, let us consider the following simple example. Example 10.1.1. We wish to find the area of the triangle with vertices (0, 2), (2, 0) and (2, 4). Consider the picture below: y mZZZZZZZ ZZZZZZ Z y =x+2 2 dy =2−x dddddd qddddddd 2 Chapter 10 : Applications of the Definite Integral x page 1 of 7 First Year Calculus c W W L Chen, 1994, 2005 Consider a dissection ∆ : 0 = x0 < x1 < . . . < xn = 2 of the interval [0, 2], and suppose that every subinterval [xi−1 , xi ] is very short. Suppose that [xi−1 , xi ] is the base of the very narrow vertical strip shown in the picture. The heights of the left hand side and right hand side of this vertical strip are respectively (xi−1 + 2) − (2 − xi−1 ) and (xi + 2) − (2 − xi ). Since xi − xi−1 is very small, the two heights are roughly the same. It follows that the area of this vertical strip is base × height = (xi − xi−1 )((ξi + 2) − (2 − ξi )), where ξi ∈ [xi−1 , xi ]. If we now consider all such strips, then the total area is the Riemann sum n (xi − xi−1 )((ξi + 2) − (2 − ξi )) i=1 of the Riemann integral 2 ((x + 2) − (2 − x)) dx. 0 It follows that the area of the triangle is 2 2 ((x + 2) − (2 − x)) dx = 0 0 2x dx = 4. Arguing in a similar way, we have the following simple result. PROPOSITION 10A. Suppose that the functions g (x) and h(x) are continuous in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that g (x) ≤ h(x) for every x ∈ [A, B ]. y y = h( x ) y = g(x ) A B x Then the area bounded by the curves y = g (x) and y = h(x) and the lines x = A and x = B is given by B (h(x) − g (x)) dx. A Chapter 10 : Applications of the Definite Integral page 2 of 7 First Year Calculus c W W L Chen, 1994, 2005 Example 10.1.2. Suppose that we wish to find the area α of the triangle with vertices (0, 2), (2, 0) and (4, 4) (the reader should try to draw a picture). We can consider the interval [0, 4] and write h(x) = 1 x+2 2 and g (x) = 2−x 2x − 4 if x ∈ [0, 2], if x ∈ [2, 4]. Note that the function g (x) is continuous in the closed interval [0, 4]. It follows from Proposition 10A that 4 2 4 α= 0 2 (h(x) − g (x)) dx = 0 (h(x) − g (x)) dx + 2 4 (h(x) − g (x)) dx dx = 0 1 x + 2 − (2 − x) 2 2 4 dx + 2 1 x + 2 − (2x − 4) 2 = 3 2 x dx + 0 2 3 6− x 2 dx = 6. Example 10.1.3. Suppose that we wish to find the area α bounded by the parabola y 2 = x + 5 and the line y = x − 1 (the reader should try to draw a picture). Note that the parabola intersects the x-axis at the point (−5, 0), and that the parabola intersects the line at the points (4, 3) and (−1, −2). We can consider the interval [−5, 4] and write h(x) = √ x+5 and g (x) = √ − x + 5 if x ∈ [−5, −1], x−1 if x ∈ [−1, 4]. Note that the function g (x) is continuous in the closed interval [−5, 4]. It follows from Proposition 10A that 4 α= −5 (h(x) − g (x)) dx = −1 −1 −5 4 (h(x) − g (x)) dx + −1 (h(x) − g (x)) dx =2 −5 √ x + 5 dx + √ 125 ( x + 5 − x + 1) dx = . 6 −1 4 Alternatively, we may interchange the roles of x and y , consider the interval [−2, 3] and write H (y ) = y + 1 It follows from Proposition 10A that 3 3 and G(y ) = y 2 − 5. α= −2 (H (y ) − G(y )) dy = −2 (6 + y − y 2 ) dy = 125 . 6 Remark. Note that in Example 10.1.3, integrating over y proves to be much simpler than integrating over x, as we do not have to break up the range of integration. This is a very important consideration. In choosing which variable to integrate, we must bear in mind two considerations. We want to minimize the number of integrations, and we also want to obtain simple definite integrals. Occasionally a little compromise may be necessary. 10.2. Volumes of Solids of Revolution In this section, we discuss the idea of rotating flat areas about a line on the same plane to produce volumes of solids of revolution. A simple example is the following. Chapter 10 : Applications of the Definite Integral page 3 of 7 First Year Calculus c W W L Chen, 1994, 2005 Example 10.2.1. Consider the triangle with vertices (0, 0), (2, 0) and (2, 1), as shown in the following diagram: y 1 r rr rr rr rr r rr rr m rr rr rr r rr rr rr rr rr r rr rr rr rr rr rr rr y = 1x 2 2 x Here we imagine that the positive z -axis is coming towards us. If we rotate the triangle about the x-axis, then we obtain a solid cone. Consider a dissection ∆ : 0 = x0 < x1 < . . . < xn = 2 of the interval [0, 2], and suppose that every subinterval [xi−1 , xi ] is very short. Suppose that [xi−1 , xi ] is the base of the very narrow vertical strip shown in the picture. The heights of the left hand side and right hand side of this vertical strip are respectively 1 xi−1 2 and 1 xi . 2 Since xi − xi−1 is very small, the two heights are roughly the same. It follows that when this vertical strip is rotated about the x-axis, we get a very thin circular slab of radius 1 ξi and volume 2 thickness × area = (xi − xi−1 )π 1 ξi 2 2 , where ξi ∈ [xi−1 , xi ]. If we now consider all such slabs, then the total volume is the Riemann sum n (xi − xi−1 )π i=1 1 ξi 2 2 of the Riemann integral 2 π 0 1 x 2 2 dx. It follows that the volume of the solid cone is 2 π 0 1 x 2 2 dx = π 4 2 x2 dx = 0 2π . 3 Arguing in a similar way, we have the following simple result. PROPOSITION 10B. Suppose that the function f (x) is continuous in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further f (x) ≥ 0 for every x ∈ [A, B ]. Then the volume obtained when the area bounded by the curve y = f (x) and the lines y = 0, x = A and x = B is rotated about the x-axis is given by B π A Chapter 10 : Applications of the Definite Integral f 2 (x) dx. page 4 of 7 First Year Calculus c W W L Chen, 1994, 2005 Note that Proposition 10B is rather restrictive, in the sense that the x-axis has to feature prominantly as part of the edge of the area in question. The following generalization is much more useful. PROPOSITION 10C. Suppose that the functions g (x) and h(x) are continuous in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that h(x) ≥ g (x) ≥ 0 for every x ∈ [A, B ]. Then the volume obtained when the area bounded by the curves y = g (x) and y = h(x) and the lines x = A and x = B is rotated about the x-axis is given by B π A (h2 (x) − g 2 (x)) dx. Proof. Suppose that G denotes the area bounded by the curve y = g (x) and the lines y = 0, x = A and x = B , and that H denotes the area bounded by the curve y = h(x) and the lines y = 0, x = A and x = B . Then the area bounded by the curves y = g (x) and y = h(x) and the lines x = A and x = B is obtained by removing G from H. The volume obtained when this is rotated about the x-axis is therefore the volume obtained when H is rotated about the x-axis minus the volume obtained when G is rotated about the x-axis. The result now follows from Proposition 10B. A natural question to ask at this point is the more general problem of rotating about a line y = y0 rather than the x-axis y = 0. We state the following general result. PROPOSITION 10D. Suppose that the functions g (x) and h(x) are continuous in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that either h(x) ≥ g (x) ≥ y0 for every x ∈ [A, B ] or h(x) ≤ g (x) ≤ y0 for every x ∈ [A, B ], where y0 ∈ R. Then the volume obtained when the area bounded by the curves y = g (x) and y = h(x) and the lines x = A and x = B is rotated about the line y = y0 is given by B π A ((h(x) − y0 )2 − (g (x) − y0 )2 ) dx. Remarks. (1) The condition that either h(x) ≥ g (x) ≥ y0 for every x ∈ [A, B ] or h(x) ≤ g (x) ≤ y0 for every x ∈ [A, B ] simply means that the curve y = g (x) is between the line y = y0 and the curve y = h(x). (2) If an area is rotated about the y -axis or a line of the form x = x0 , then we have to integrate with respect to y . Example 10.2.2. Let us try to find the volume of a sphere of radius r. To do this, we shall rotate a half-disc of radius r. More precisely, consider the half-disc on the upper half-plane centred at the origin and of radius r. In other words, we consider the area bounded by the circle x2 + y 2 = r2 , with y ≥ 0. In the notation of Proposition 10B, we consider the function f (x) = It follows that the volume in question is r r2 − x2 , where x ∈ [−r, r]. π −r Chapter 10 : Applications of the Definite Integral (r2 − x2 ) dx = 4πr3 . 3 page 5 of 7 First Year Calculus c W W L Chen, 1994, 2005 Example 10.2.3. Consider the triangle with vertices (0, 1), (2, −1) and (3, 2). We wish to find the volume µ when this triangle is rotated about the line x = −1. We have the following picture: ;; ;; ;; ;; ;; ;; ;; x=−1 y ;; ;; ;; ;; ;; ;; ;; ;; ;; ; ;; kk kkk kkk kkk kkk kkk kkk kkk m kkk kkk kkk 7kk x = 3y − 3 7 1 77 7 7 7 7 7 7 7 mZZZZZZZ 7 7 ZZZZZZ 7 7 Z 7 7 7 x = 1y + 7 7 x 7 3 3 2 3 7 7 7 7 7m 7 7 7 7 7 7 7 −1 2 x=1−y We can consider the interval [−1, 2] and write H (y ) = 1 7 y+ 3 3 and G(y ) = 1−y 3y − 3 if y ∈ [−1, 1], if y ∈ [1, 2]. Note here that the function G(y ) is continuous in the closed interval [−1, 2], and that −1 ≤ G(y ) ≤ H (y ) in [−1, 2]. It follows from Proposition 10D that 2 µ=π −1 1 ((H (y ) + 1)2 − (G(y ) + 1)2 ) dy 2 =π −1 1 ((H (y ) + 1)2 − (G(y ) + 1)2 ) dy + π 1 ((H (y ) + 1)2 − (G(y ) + 1)2 ) dy 2 =π −1 1 10 y+ 3 3 2 − (2 − y )2 dy + π 1 1 10 y+ 3 3 2 − (3y − 2)2 dy = 64π . 3 Problems for Chapter 10 1. Find the area enclosed by the curves y = x2 and y = x4 . 2. Find the area enclosed by the four lines x + y = 1, x + y = 5, x − 3y = 1 and x − 3y = −3. 3. Find the area bounded by the curve x = 4y − 4y 2 and the lines x − y = 3, y = 0 and y = 1. 4. Let R be the region in the first quadrant bounded by the curve y = sin−1 x, the x-axis and the line x = 1/2. Give a sketch of the region R and determine its area. 5. Suppose that α is a positive real number and n > 1 is an integer. 2x a) Find the area Sn bounded by the curve y = 2 and the x-axis between the lines x = 1 (x + 1)α and x = n. b) Find all values of α for which the limit of Sn is finite as n → +∞. 6. Use integration to show that the volume of a sphere of radius R is 4 πR3 . 3 Chapter 10 : Applications of the Definite Integral page 6 of 7 First Year Calculus c W W L Chen, 1994, 2005 7. For each of the lines below, find the volume obtained when the area bounded by the parabolas y = 1 − x2 and y = 3 − 3x2 is rotated about the line: a) y = 0 b) y = −2 c) y = 4 8. a) b) c) d) Sketch the curves y = x and y = ex on the same coordinate plane. Find the area bounded by the two curves in part (a), the y -axis and the line x = 1. Find the volume created when the area in part (b) is rotated about the x-axis. Find the volume created when the area in part (b) is rotated about the y -axis. 9. Let S be the region bounded by the curve y = cos 2x for 0 ≤ x ≤ π/4, the x-axis and the y -axis. Determine the volume generated when the region S is rotated about the line x = −1. Chapter 10 : Applications of the Definite Integral page 7 of 7 ...
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