To determine The volume of the solid that is common to the three cylinders. Explanation Given: The coordinate axes form the axes of the three right circular cylinders with radius 1. Calculation: The calculation of volume of the solid that is common to the three cylinders is as follows. Due to symmetry, the volume of the region of the intersection of the three right circular cylinders each of radius 1 is made up of eight identical pieces, one in each octant. Let one consider the piece of the volume in the first octant. A particle moving through this region parallel to the positive y -axis will start moving in any direction on x z -plane within the unit circle x 2 + z 2 = 1 . And the particle that started from parallel to the positive y -axis would either end on the unit radius cylinder that runs parallel to the x -axis ( y 2 + z 2 = 1 ) or that runs parallel to the z -axis ( x 2 + y 2 = 1 ) as shown below in Figure 1. Suppose the particle starts at a point on the x z -plane for which x < z , by using the inequality property 1 − z 2 < 1 − x 2 and the particle ends on the unit cylinder y 2 + z 2 = 1 . Let us assume in above case R be the region which volume given by triple integral ∭ R d y d z d x . The boundaries of the region R are as follows. R = { 0 < x < 2 2 ; 0 < y < 1 − z 2 , and x < z < 1 − x 2 } Similarly, the particle starts at a point on the x z -plane for which z < x , by using the inequality property 1 − x 2 < 1 − z 2 and the particle ends on the unit cylinder x 2 + y 2 = 1 . Again assume in above case R * be the region which volume given by triple integral ∭ R * d y d x d z . The boundaries of the region R * are as follows. R * = { 0 < z < 2 2 ; 0 < y < 1 − x 2 , and z < x < 1 − z 2 } Suppose V be the total volume of the three interesting cylinders which is 8 times of the volume of one piece in first octant. V = 8 ( ∭ R d y d z d x + ∭ R * d y d x d z ) = 8 ∫ x = 0 x = 2 2 ∫ z = x 1 − x 2 ∫ y = 0 1 − z 2 1 d y d z d x + 8 ∫ z = 0 z = 2 2 ∫ x = z 1 − z 2 ∫ y = 0 1 − x 2 1 d y d x d z = 8 ∫ x = 0 x = 2 2 ∫ z = x 1 − x 2 ( y ) | 0 1 − z 2 d z d x + 8 ∫ z = 0 z = 2 2 ∫ x = z 1 − z 2 ( y ) | 0 1 − x 2 d x d z = 8 ∫ x = 0 x = 2 2 ∫ z = x 1 − x 2 1 − z 2 d z d x + 8 ∫ z = 0 z = 2 2 ∫ x = z 1 − z 2 1 − x 2 d x d z Simplified further as, V = 8 ∫ x = 0 x = 2 2 ∫ z = x 1 − x 2 1 − z 2 d z d x + 8 ∫ z = 0 z = 2 2 ∫ x = z 1 − z 2 1 − x 2 d x d z = 8 ∫ x = 0 x = 2 2 ( z 2 1 − z 2 + 1 2 sin − 1 z ) |

Calculus: Early Transcendentals (3rd Edition)

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Chapter 16.4, Problem 64E

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The volume of the solid that is common to the three cylinders.

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Chapter 16.4, Problem 63E

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Chapter 16 Solutions

Calculus: Early Transcendentals (3rd Edition)

Ch. 16.1 - Explain why the sum for the volume is an...Ch. 16.1 - Consider the integral 3412f(x,y)dxdy. Give the...Ch. 16.1 - Prob. 3QC Ch. 16.1 - 1. Write an iterated integral that gives the...Ch. 16.1 - Write an iterated integral that gives the volume...Ch. 16.1 - Write two iterated integrals that equal Rf(x,y)dA,...Ch. 16.1 - Consider the integral 1311(2y2+xy)dydx. State the...Ch. 16.1 - Region R = {(x, y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 6} is...Ch. 16.1 - Pictures of solids Draw the solid whose volume is...Ch. 16.1 - Iterated integrals Evaluate the following iterated...

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