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.
bro whats the base equation again, something like pi int .... of the derivative squared?
idk if that matters
i have my integral calculus last semester.
y=+or- sqrtx , y=x/2 to make it simpler
come on guys, this isnt that hard.
wait for a while, i'm solving it.
what is really the given? it is lacking.
"find the volume of the solid obtained by rotating the regions bounded by the curves about the specified line.
you have to find the point where the two curves meet, and then solve the integral from 0 to that point
they meet at (0,0)
thats one yes, and then they cross again
the answer is
8pi over 3
where do the two curves cross?
the limits are 0 and 2.
just substitute x=y^2 to x=2y that will become, y^2-2y=0 y(y-2)=0 y=0 y=2
okay, let me double check now
i had a mistake, the answer must be 10pi/3
almost have it
okay, i have the solution but i have a hard time encoding it.
whats (2^1.5)/(2/3) -1
i dont have my calculator :/
okay thats what i got... where did you get pi from?
im looking in the back of the book for the answer to see whos right.
2pi integral of (x^1/2-x/2)dx 2pi (x^3/2)/(3/2)-x^2/4, substitute x=2 2pi(2^3/2)(2/3)-2^2/4 2pi(4(2/3)-1) 2pi(8/3 - 1) 2 pi (5/3) (10pi)/3
okay i did it waaaay wrong haha
back of the book says 64pi/15
is it correct now?
is it the answer on the book?
64pi/15, so its close
okay im too tired for this, thanks for the help
you are welcome, actually i can explain it, i think i just miss some terms in my solution. by the way, do you have skype account?