We need to verify the formula ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = 1 6 ( x 1 − x 2 ) 3 . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = π 6 B H 2 .

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Answer 1

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Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

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Answer 2

Question

Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

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Answer 3

Question

Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

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Answer 4

Question

Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

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Answer 5

Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

Answer 6

Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

Answer 7

Chapter 6.3, Problem 65E

To determine

We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 6.1 - Prob. 1PQ Ch. 6.1 - Prob. 2PQ Ch. 6.1 - Prob. 3PQ Ch. 6.1 - Prob. 4PQ Ch. 6.1 - Prob. 5PQ Ch. 6.1 - Prob. 6PQ Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E

We need to verify the formula ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = 1 6 ( x 1 − x 2 ) 3 . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = π 6 B H 2 .

Solution Summary: The author proves that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = 6 B H 2.

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We need to verify the formula $\int_{x_{1}}^{x_{2}} (x - x_{1}) (x - x_{2}) d x = \frac{1}{6} {(x_{1} - x_{2})}^{3}$ . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume $V = \frac{π}{6} B H^{2}$ .

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Chapter 6 Solutions

We need to verify the formula ∫ x 1 x 2 ( x − x 1 ) ( x − x 2 ) d x = 1 6 ( x 1 − x 2 ) 3 . Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = π 6 B H 2 .

Solution Summary: The author proves that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = 6 B H 2.

Videos