Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find the volume generated by revolution of the region y = x and y = 3-x², around the y-axis

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

|dw:1349809459606:dw|
just this part ?
i think i need to consider the negative x values too

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1349809625381:dw|
|dw:1349809619079:dw|
this part rotated around the y-axis
ok so intersection of y=x and y=3-x^2 is 3 -x^2 =x x^2 +x -3 = 0 x = 0.5(-1-sqrt(13)) x = 0.5(sqrt(13)-1)
so i think we should separate into positive x and negative x
for 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(y-3))^2dy from y = 0.5(sqrt(13)-1) to y= 3
agree so far?
yes
wait .. i think that we dont need the positive x part at all
since the revolution of the negative part will cover it
i was thinking in use \[\int\limits_{0.5(-1-\sqrt(13))}^{0.5(\sqrt(13)-1)}2\pi x (-x² -x +3) dx\]
so i would go only for the negative part.. there are two parts : integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0 and integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3
them after i calculate this two parts i sum them?
wait 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"
i would only add those 2 : integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0 and integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3
yeah, after the rotation it will cover the positive part
ok a got it! thank you!
so if we add the positive part we get an extra for the answer and its bad..
yeah... that question is fun rsrsr
lol so you agree with : integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0 and integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3 ?
only one thing is strange, sholdn't it be integral of pi*((-sqrt(y-3))-y)^2*dy from y= 0.5(-1-sqrt(13)) to y = 0 and integral of pi*(sqrt(y-3))^2*dy from y = 0 to y = 3 ?
(a-b)^2 = (b-a)^2
because the function 3-x² is in far of y = x in relation to the y-axis?
(a-b)^2 = (b-a)^2 !!
and anyway it's better (for me at least) to think of subtracting the small number from the high number
ohh ok i got it!
thank you!!
;)
oh wait!!
there is something wrong there!
integral of pi*(y-(-sqrt(y-3)))^2*dy from y= 0.5(-1-sqrt(13)) to y = 0 i should square them both -> y-(-sqrt(y-3)) we should do : integral of pi*[(sqrt(y-3))^2 - y^2]*dy from y= 0.5(-1-sqrt(13)) to y = 0
sorry for that.. so long havent done such things
because we like subtracting the inner disc from the larger disc

Not the answer you are looking for?

Search for more explanations.

Ask your own question