revolve the region in first quadrant bounded by y = x^2 , y = x , about line x = 2, using washer method

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

revolve the region in first quadrant bounded by y = x^2 , y = x , about line x = 2, using washer method

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\pi \int\limits_{0}^{4}(2-x)^{2}-(2-x ^{2})^{2}\]
thats wrong
noooooooooo

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

srry its been a long time.
the region is from (0,0) to (1,1)
oh sorry sorry forgot about the bounds. assumed that x22 was also a bound
also youre using dy , not x
ok nvm i rambl.ed
now you will die
ure supposed to use dx?
dy
i was using DY
since its horizontal integration
just the wrong values
no its wrong in other ways!!!
ok lets do it again: \[\pi \int\limits_{0}^{1} (2-x)^{2}-(2-x ^{2})^{2}\]
this HASSSSS TO BE RIGHT
eh eh?
no
its pi* integral ( y - 2)^2 - (sqrt y - 2)^2

Not the answer you are looking for?

Search for more explanations.

Ask your own question