A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Find the volume generated by revolution of the region y = x and y = 3x², around the yaxis
anonymous
 4 years ago
Find the volume generated by revolution of the region y = x and y = 3x², around the yaxis

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349809459606:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think i need to consider the negative x values too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349809625381:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349809619079:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this part rotated around the yaxis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so intersection of y=x and y=3x^2 is 3 x^2 =x x^2 +x 3 = 0 x = 0.5(1sqrt(13)) x = 0.5(sqrt(13)1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i think we should separate into positive x and negative x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for the positive x there are two parts : integral of pi*(y)^2dy from y =0 to y = 0.5(sqrt(13)1) and integral of pi*(sqrt(y3))^2dy from y = 0.5(sqrt(13)1) to y= 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait .. i think that we dont need the positive x part at all

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since the revolution of the negative part will cover it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i was thinking in use \[\int\limits_{0.5(1\sqrt(13))}^{0.5(\sqrt(13)1)}2\pi x (x² x +3) dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i would go only for the negative part.. there are two parts : integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0them after i calculate this two parts i sum them?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait but dont you agree with: " i think that we dont need the positive x part at all since the revolution of the negative part will cover it"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i would only add those 2 : integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, after the rotation it will cover the positive part

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok a got it! thank you!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so if we add the positive part we get an extra for the answer and its bad..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah... that question is fun rsrsr

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol so you agree with : integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0only one thing is strange, sholdn't it be integral of pi*((sqrt(y3))y)^2*dy from y= 0.5(1sqrt(13)) to y = 0 and integral of pi*(sqrt(y3))^2*dy from y = 0 to y = 3 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because the function 3x² is in far of y = x in relation to the yaxis?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and anyway it's better (for me at least) to think of subtracting the small number from the high number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is something wrong there!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0integral of pi*(y(sqrt(y3)))^2*dy from y= 0.5(1sqrt(13)) to y = 0 i should square them both > y(sqrt(y3)) we should do : integral of pi*[(sqrt(y3))^2  y^2]*dy from y= 0.5(1sqrt(13)) to y = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry for that.. so long havent done such things

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because we like subtracting the inner disc from the larger disc
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.