Question

Notify Me Bookmark

For the region in the first quadrant bounded by y = 1 − x2, the x-axis, and y-axis, determine which of the following is greater: the volume of the solid generated when the region is revolved about the x-axis or about the y-axis. When the region is revolved about the x-axis, the volume is . (Type an exact answer, using π as needed.) When the region is revolved about the y-axis, the volume is. (Type an exact answer, using π as needed.) Which volume is greater? A. The volume about the x-axis is greater. B. The volumes are equal. C. The volume about the y-axis is greater.

For the region in the first quadrant bounded by y = 1 − x2, the x-axis, and y-axis, determine which of the following is greater: the volume of the solid generated when the region is revolved about the x-axis or about the y-axis. When the region is revolved about the x-axis, the volume is . (Type an exact answer, using π as needed.) When the region is revolved about the y-axis, the volume is. (Type an exact answer, using π as needed.) Which volume is greater? A. The volume about the x-axis is greater. B. The volumes are equal. C. The volume about the y-axis is greater.

Image text
For the region in the first quadrant bounded by y = 1 − x 2 , the x -axis, and y -axis, determine which of the following is greater: the volume of the solid generated when the region is revolved about the x -axis or about the y -axis.
When the region is revolved about the x -axis, the volume is ◻ ◻ . (Type an exact answer, using π as needed.) When the region is revolved about the y -axis, the volume is ◻ . (Type an exact answer, using π as needed.) Which volume is greater? A. The volume about the x -axis is greater. B. The volumes are equal. C. The volume about the y-axis is greater.

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

Find the volume of the solid generated by revolving the region about the given line. The region in the first quadrant bounded above by the line y = 2/3, below by the curve y = csc xcot x, and on the right by the line x = π/2, about the line y = 2/3. Set up the integral that gives the volume of the solid. (Type exact answers, using π as needed.) The volume of the solid generated is cubic units. (Type an exact answer, using π as needed.)
Solution
0 0

Find the centroid (center of mass) of the following thin plate, assuming constant density. Use symmetry when possible to simplify the calculations. The region in the first quadrant bounded by x2 + y2 = 289 Make a sketch of the region. Choose the correct answer below. A. B. C. D. Set up the double integral that gives My, the plate's first moment about the y -axis. Use increasing limits of integration. Assume a density of 1. My = ∫()dydx (Type exact answers.) A. This integral is not necessary because the given function is symmetric about the y-axis. B. This integral is not necessary because the given function is symmetric about the line y = x. C. This integral is not necessary because the given function is symmetric about the x-axis. Dhis integral is necessary; Mx = ∬(dydx (Type exact answers.) The coordinates of the centroid are (x¯, y¯) = (Type an ordered pair. Type an exact answer, using π as needed.) Plot the centroid on the sketch of the region. Choose the correct answer below. A. B. C. D.
Solution
0 0

Find the area of the surface generated when the given curve is revolved about the x-axis. y = 4x+6 on [0, 3] The area of the generated surface is square units. (Type an exact answer, using π as needed.)
Solution
0 0