Posted by Kira on Monday, March 5, 2007 at 7:32pm.
Hi. I have a few calculus problems that I have to do using the washers and shell method to find out what the volume of the graph rotated is. I'll post one question at a time with my work. The problem is, I should be getting the same answer for the problem with the shell method and the washer method, but I'm getting two different answers. If someone could check my work, it would be GREATLY appreciated! And please keep explanations in as easy a form as you can. I'm having a really hard time understanding these concepts. THANKS!!!
1. Find the volume of the solid that results when the region bounded by y=x and y=x^2 from x=0 and x=1, is resolved about the x axis. Apply both the shell and washer method to solve.
2pi(INT from 0 to 1)y*sqrt(y)dy
2pi(INT from 0 to 1)y^3/2dy
2pi[2/5y^5/2]from 0 to 1
pi(INT from 0 to 1)x^2-x^4 dx
pi[1/3x^3 - 1/5x^5]from 0 to 1
pi[5/15 - 3/15]
I don't know how to do shell method, but, I'd say do it on your calculator and see which answer is right and go from there
Your second answer is right. In the first one, the hight is y but the radius is (sqrt(y)-y). Look at a graph to see why. So:
2pi(INT from 0 to 1)y(sqrt(y)-y)dy
2pi(INT from 0 to 1)[y^(3/2)-y^2]dy
Answer This Question
More Related Questions
- Math - Hi, I have to do a project on the Shell Method in my Geometry class. I ...
- Calculus - Volume By Integration - Find the volume of the solid generated by ...
- Calculus - Find volume of cylinder - There is a cylindrical tank lying ...
- calculus - The volume of the solid obtained by rotating the region enclosed by ...
- calculus - The volume of the solid obtained by rotating the region enclosed by...
- please help calculus - find the volume of the solid formed by revolving the ...
- calculus - the region bounded by the graph f(x)=x(2-x) and the x axis is ...
- Calculus - *Note I reposted this question as I changed the subject** The Region...
- Calculus - Find the volume using the shell method - about the x-axis x=y^2/3
- Urgent Math - Suppose the area under y = -x^2+1 between x = 0 and x = 1 is ...