A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Find the volume of the solid generated from revolving y = sqrt(4x^2) around the x axis.
anonymous
 4 years ago
Find the volume of the solid generated from revolving y = sqrt(4x^2) around the x axis.

This Question is Closed

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0amistre, please help me asap.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i think this would be infinite without bounds

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0Ben he will be back dont worry. i have 3mins. ;(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well it has zeros at 2 and 2 i think. i ended up getting an answer i just dont know how to check it. i got 32pi/3

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1to check, wolframalpha does a wonderful job

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.12pi is what the wolf gives us http://www.wolframalpha.com/input/?i=integrate+sqrt%284x%5E2%29+from+2+to+2

saifoo.khan
 4 years ago
Best ResponseYou've already chosen the best response.0Amistre, another question please?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if we do what is commonly refered to as a disc method we add up the areas of a bunch of circles; A = pi r^2, where r = our given function.\[pi\int _{0}^{2}(\sqrt{4x^2})^2dx\] \[pi\int _{0}^{2}(4x^2)dx=pi(4x\frac{1}{3}x^3)\] \[pi(4(2)\frac{1}{3}2^3)0\] \[pi(8\frac{8}{3})\]i see the conundrum alright :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is wrong with that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and souldent it be 2pi on the outside of the integral? you would either have to do it from 2 to 2 or you would need to do 2pi integral from 0 to 2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1nah, the area of each circle is: pi [f(x)]^2 and we add all those up from 2 to 2, or just from 0 to 2 and double it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1(sqrt(a))^2 = a so we can forgo the sqrt in the integration int 4  x^2 4 ups to 4x x^2 ups to x^3/3 4x  x^3/3 at 0 = 0 at 2 = 8 8/3 = 16/3 and not to forget the pi http://www.wolframalpha.com/input/?i=integrate+pi%28sqrt%284x%5E2%29%29%5E2+from+0+to+2 now the wolf agrees

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1so yeah, 32/3 pi should be it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i see, i entered it wrong at the start of all this .... it sucks getting old lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha ok cool thank you
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.