anonymous
  • anonymous
how do you find the volume of a solid made by rotating y=2(x-1)^(1/2) and y=x-1 about the line x=-1 using the shell method?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Ok well I'm sure you have the formula for the shell method. So all you need is the radius, height, and the boundary points for your cylinders. The radius should be x+1, the boundary for x is 1 to 5. The height is going to be the upper graph minus the lower. Can you do the rest?
anonymous
  • anonymous
It'd be tough to explain through open study since drawings are necessary to understand it. I'd recommend going to this http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx to help
anonymous
  • anonymous
you can use an online whiteboard, try twiddla.com

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I've tried it.. I got the height, but I can't get the right radius
anonymous
  • anonymous
I tried it using disc and got it, but can't figure it out using shell
anonymous
  • anonymous
What was your integral? was it \[\int\limits_{1}^{5} 2\pi (x+1)[2\sqrt{x-1} - (x - 1)] dx?\] The radius is (x + 1). Normally when you revolve around the y-axis (the line x = 0) your radius is just x. But now you've shifted the axis left one to get x = -1, so the radius increased by one. I think example 3 in the link I gave you is close to your problem.
anonymous
  • anonymous
oh that makes sense.

Looking for something else?

Not the answer you are looking for? Search for more explanations.