(a) Explanation Given: A radius of spherical object is 25 centimetres. A hole of radius 4 centimetres is drilled through centre of the object. Formula Used: The volume can be found by double integrating the surface within the limits. V = ∫ ∫ ( f ( x , y ) ) d A The equation of sphere is x 2 + y 2 + z 2 = r 2 . The equation of circle is x 2 + y 2 = r 2 . Integration formula ∫ x n d x = x n + 1 n + 1 . Calculation: As provided the object is sphere of radius 25 centimetres and a hole of radius 4 centimetres is drilled through centre of the object. The region is x 2 + y 2 + z 2 = 25 2 which is a sphere and a hole of radius 4 centimetres is drilled through centre of the object that is x 2 + y 2 + z 2 = 4 2 . Then, z = f ( x , y ) = 25 2 − x 2 − y 2 Then the net region appears in x-y plane in the first quadrant is as: For the region, the limits shall be converted to polar points for both x and y . x = r cos θ and y = r sin θ , which implies to, {by using sin 2 x + cos 2 x = 1 } x 2 + y 2 = r 2 ( cos 2 x + sin 2 x ) = r 2 Then the region is bounded by 4 ≤ r ≤ 25 and 0 ≤ θ ≤ π 2 . The equation of the region is: z = 25 2 − x 2 − y 2 = 25 2 − ( x 2 + y 2 ) = 25 2 − r 2 Substitute the above equation in the formula of volume. The surface is multiplied by 8 to account for all the hemisphere V = 8 ∫ 0 π 2 ∫ 4 25 25 2 − r 2 r d r d θ Take u = 25 2 − r 2 , d u = − 2 r d r and apply formula ∫ x n d x = x n + 1 n + 1 (b) To determine To calculate: The outer surface area of an object which is a spherical object of radius 25 centimetres with a hole of radius 4 centimetres is drilled through centre of the object.

6th Edition

ISBN: 9780100475557

Author: Edwards

Publisher: YUZU

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Question

Chapter 14.5, Problem 33E

(a)

To determine

To calculate: The volume of the object where the radius of spherical object is 25 centimetres and a hole of radius 4 centimetres is drilled through the centre of the object.

(b)

To determine

To calculate: The outer surface area of an object which is a spherical object of radius 25 centimetres with a hole of radius 4 centimetres is drilled through centre of the object.

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Chapter 14.5, Problem 32E

Chapter 14.5, Problem 34E

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Chapter 14 Solutions

EBK CALCULUS: EARLY TRANSCENDENTAL FUNC

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - Prob. 1E Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy

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The volume of the object where the radius of spherical object is 25 centimetres and a hole of radius 4 centimetres is drilled through the centre of the object.

Solution Summary: The author explains how the volume of the object can be found by double integrating the surface within the limits.