Explanation Given: The equation of the cylinder is x 2 + y 2 = x And the equation of paraboloid is z = 1 − x 2 − y 2 Calculation: Let’s take r = 1 a t z = 0 x = r cos θ , d x = − r sin θ y = r sin θ , d y = r cos θ r 2 = r cos θ r = cos θ a t z = 1 − x 2 − y 2 So, z = 1 − r 2 cos 2 θ − r 2 sin 2 θ Then the integral will be V = ∫ 0 2 π ∫ 1 cos θ ∫ 0 1 − r 2 cos 2 θ − r 2 sin 2 θ r d z d r d θ V = ∫ 0 2 π ∫ 1 cos θ ( 1 − r 2 cos 2 θ − r 2 sin 2 θ ) r d r d θ V= ∫ 0 2 π ( r − r 4 cos 2 θ 4 − r 4 sin 2 θ 4 ) 1 cos θ d θ V =

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ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 12.3, Problem 55PS

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Chapter 12.3, Problem 54PS

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This question builds on an earlier problem. The randomized numbers may have changed, but have your work for the previous problem available to help with this one. A 4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The wheel rotates counterclockwise at 1.5 rev/sec. At some point, the rod will be tangent to the circle as shown in the third picture. A B A B at some instant, the piston will be tangent to the circle (a) Express the x and y coordinates of point A as functions of t: x= 2 cos(3πt) and y= 2 sin(3t) (b) Write a formula for the slope of the tangent line to the circle at the point A at time t seconds: -cot(3πt) sin(3лt) (c) Express the x-coordinate of the right end of the rod at point B as a function of t: 2 cos(3πt) +411- 4 -2 sin (3лt) (d)…

5. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.AE.003. y y= ex² 0 Video Example x EXAMPLE 3 (a) Use the Midpoint Rule with n = 10 to approximate the integral कर L'ex² dx. (b) Give an upper bound for the error involved in this approximation. SOLUTION 8+2 1 L'ex² d (a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.) dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)] 0.1 [0.0025 +0.0225 + + e0.0625 + 0.1225 e0.3025 + e0.4225 + e0.2025 + + e0.5625 €0.7225 +0.9025] The figure illustrates this approximation. (b) Since f(x) = ex², we have f'(x) = 0 ≤ f'(x) = < 6e. ASK YOUR TEACHER and f'(x) = Also, since 0 ≤ x ≤ 1 we have x² ≤ and so Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final answer to five decimal places.) 6e(1)3 e 24( = ≈

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Chapter 12 Solutions

Calculus

Ch. 12.1 - Prob. 1PS Ch. 12.1 - Prob. 2PS Ch. 12.1 - Prob. 3PS Ch. 12.1 - Prob. 4PS Ch. 12.1 - Prob. 5PS Ch. 12.1 - Prob. 6PS Ch. 12.1 - Prob. 7PS Ch. 12.1 - Prob. 8PS Ch. 12.1 - Prob. 9PS Ch. 12.1 - Prob. 10PS

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Solution Summary: The author explains that the integral is V=-94pi.

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