Question

Notify Me Bookmark

The region R in the first quadrant of the xy-plane is bounded by the curves y = −3x2 + 15x + 18, x = 0 and y = 0. A solid S is formed by rotating R about the y-axis: The (exact) volume of S is: Preview The region R in the first quadrant of the xy-plane is bounded by the curves y = −6sin⁡(x), x = π, x = 2π and y = 0. A solid S is formed by rotating R about the y-axis: The volume of S is: Preview

The region R in the first quadrant of the xy-plane is bounded by the curves y = −3x2 + 15x + 18, x = 0 and y = 0. A solid S is formed by rotating R about the y-axis: The (exact) volume of S is: Preview The region R in the first quadrant of the xy-plane is bounded by the curves y = −6sin⁡(x), x = π, x = 2π and y = 0. A solid S is formed by rotating R about the y-axis: The volume of S is: Preview

Image text
The region R in the first quadrant of the x y -plane is bounded by the curves y = − 3 x 2 + 15 x + 18 , x = 0 and y = 0 . A solid S is formed by rotating R about the y -axis:
The (exact) volume of S is: Preview The region R in the first quadrant of the x y -plane is bounded by the curves y = − 6 sin ⁡ ( x ) , x = π , x = 2 π and y = 0 . A solid S is formed by rotating R about the y -axis:
The volume of S is: ◻ Preview

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

Let R be the finite region bounded by curves y = 1/x and 3x + 3y = 10. A solid S is obtained by rotating the region R about the y-axis. Which of the following integral represents the volume of S ? Select one: a. π ∫ 3 1/3 1/x2 - 10-3x 3 2 dx b. π ∫ 3 1/3 1/y2 - 10-3y 3 2 dy c. π ∫ 3 1/3 10-3y 3 2 - 1/y2dy d. π ∫ 3 1/3 10-3x 3 2 - 1/x2 dx

Explain how you got to the correct answer. THANK YOU
Solution
1 0

Consider the region R bounded by the graphs of y = (x) > 0, x = a > 0, x = b > a, and y = 0. (a) Suppose f(x) = 4x2 − x4 with a = 1, b = 2. Calculate the volume of the solid formed by revolving R about the x-axis. (b) Do the same for the solid formed by revolving R about the y-axis
Solution
0 0

A computer algebra system is recommended. Let R be the region in the first quadrant bounded by the y-axis and the graphs of y = x2 + 3 and y = 8 − ex. (a) Find the area of the region R. (Round your answer to three decimal places.) (b) Find the volume of the solid that results when R is rotated about the x-axis. (Round your answer to three decimal places.) (c) The solid S has base R. Each cross-section perpendicular to the x-axis is a semicircle whose diameter lies in R. Find the volume of the solid S. (Round your answer to three decimal places ∫0 k(8−ex)−(x2+3)dx = ∫ka(8−ex)−(x2+3)dx ∫0 k(8−ex)dx = ∫0 k(x2+3)dx π∫0 k(8−ex)2−(x2+3)2 dx = π∫ka(8−ex)2−(x2+3)2 dx ∫0 a(8−ex)−(x2+3)dx = ∫0 k(8−ex)−(x2+3)dx ∫ka(8−ex)dx = ∫ka(x2+3)dx
Solution
0 0