Explanation Formula used: “If the function f ( x ) ≥ 0 is continuously differentiable on [ a , b ] , the area of the surface generated by revolving the graph of y = f ( x ) about the x axis is S = ∫ a b 2 π y 1 + ( d y d x ) 2 d x ”. Calculation: The given function is y = r 2 − x 2 and the area of the AB curve in interval between a to a + h on the x axis. So, the surface area of the curve is calculated as: S = ∫ a a + h 2 π y 1 + ( d y d x ) 2 d x = ∫ a a + h 2 π r 2 − x 2 1 + ( d ( r 2 − x 2 ) d x ) 2 d x ⋅ ⋅ ⋅ (1) First find out the value of d y d x . So, differentiate the function y with respect to x . d d x y = d d x ( r 2 − x 2 ) = 1 2 × − 2 x r 2 − x 2 = − x r 2 − x 2 The value of d y d x substitute in the equation (1)

14th Edition

ISBN: 9780136208013

Author: Hass

Publisher: PEARSON

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Chapter 6.4, Problem 28E

To determine

To show: That the area swept out by AB depend upon the length of the given interval.

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Chapter 6.4, Problem 27E

Chapter 6.4, Problem 29E

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Question 4 Find an equation of (a) The plane through the point (2, 0, 1) and perpendicular to the line x = y=2t, z=3+4t. 3t, (b) The plane through the point (3, −2, 8) and parallel to the plane z = x+y. (c) The plane that contains the line x = parallel to the plane 5x + 2y + z = 1. 1+t, y2t, z = 43t and is (d) The plane that passes through the point (1,2,3) and contains the line x = 3t, y=1+t, and z = 2 – t. (e) The plane that contains the lines L₁ : x = 1 + t, y = 1 − t, z = = L2 x 2s, y = s, z = 2. 2t and

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