anonymous
  • anonymous
using shell method x=y√x−.2yy13findx−axis cani find the help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
i think that needs an edit
anonymous
  • anonymous
ok
amistre64
  • amistre64
its like your trying to speak calculus to me .. what is it lassie? timmys in the well?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

dumbcow
  • dumbcow
:)
anonymous
  • anonymous
use shell method to find the volume generated by revolving the region bounded by x=y√x−.2yy13findx−axis
dumbcow
  • dumbcow
shell method: \[\rightarrow 2\pi \int\limits_{?}^{?}radius*height\]
amistre64
  • amistre64
getting warmer
amistre64
  • amistre64
what does "x=y√x−.2yy13findx−axis" mean?
anonymous
  • anonymous
no it is x= 2sqr(y) and x=0.2y and y=13
anonymous
  • anonymous
find x-axis
amistre64
  • amistre64
|dw:1333251311787:dw|
amistre64
  • amistre64
im assuming its something like this then
anonymous
  • anonymous
ok
amistre64
  • amistre64
we integrate the shell method but subtract one function from the other to remove the "odd" space thats a little up from the y axis
dumbcow
  • dumbcow
what do you mean "find x-axis" ? don;t you mean revolve the region around the x-axis
amistre64
  • amistre64
\[2pi\int _{0}^{13}y(\sqrt{y}-.2y)dy\]
amistre64
  • amistre64
dropped a 2
amistre64
  • amistre64
spose to be a 2sqrt(y)
anonymous
  • anonymous
\[i get as v=\int\limits_{0}^{13} 2\pi(y)(2\sqrt{y}+.2y)dy\]
amistre64
  • amistre64
yes, that it
anonymous
  • anonymous
this is wat i got but my answer was wrong as =3983.14
amistre64
  • amistre64
or\[2pi\int(2y^{3/2}-.2y^2)dy\]
amistre64
  • amistre64
i see you have a + in there, is that a typo or a mistake?
anonymous
  • anonymous
yes but .2y actually is negative in the orgin function
amistre64
  • amistre64
well, \[2pi(5y^{5/2}-\frac{2}{30}y^3);[0,13]\] well, \[2pi(5(13)^{5/2}-\frac{2}{30}(13)^3)\]
anonymous
  • anonymous
ok
amistre64
  • amistre64
i get abt 5947pi
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=integrate+2pi%28y%29%282sqrt%28y%29-.2y%29+from+0+to+13 or at least i thought i did, musta hit a wrong button tho
amistre64
  • amistre64
i see it, i went 5/2 instead of 2/5 when i ntegrated the coeef out
anonymous
  • anonymous
am trying to put it now cause for the first one give deffer number then walfarm
amistre64
  • amistre64
\[2pi(\frac{4}{5}y^{5/2}-\frac{y^3 }{15})\] was what i shoulda had
anonymous
  • anonymous
so round nearest to tenth
anonymous
  • anonymous
yes 299.67 to 299.674
amistre64
  • amistre64
looks good to me
anonymous
  • anonymous
the 299.674
anonymous
  • anonymous
it still wrong i dont

Looking for something else?

Not the answer you are looking for? Search for more explanations.