anonymous
  • anonymous
Set up the double integral for the volume bounded between the surface z=xy^2 and the plane z-3y-x. 0
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
|dw:1352988377508:dw|
amistre64
  • amistre64
hmmm, does the plane the the surface intersect ?
TuringTest
  • TuringTest
Shouldn't the formula for the plane have an equals somewhere?

Looking for something else?

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

More answers

amistre64
  • amistre64
1 Attachment
amistre64
  • amistre64
|dw:1352988979091:dw| pfft, i cant get a clear idea of the shape
amistre64
  • amistre64
z=xy^2 and the plane z-3y-x. 0
amistre64
  • amistre64
\[\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~\int_{3y+x}^{xy^2} ~dz~dy~dx\] maybe
amistre64
  • amistre64
\[\int_{-3\sqrt3/2}^{3\sqrt3/2}~\int_{0}^{\sqrt3}~{xy^2}-{(3y+x)}~dy~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~{\frac13x(\sqrt3)^3}-{(\frac1233+x\sqrt3)}~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~{x\sqrt3}-{\frac92-x\sqrt3}~dx\] \[\int_{-3\sqrt3/2}^{3\sqrt3/2}~-\frac92~dx\] its either zero or i might have my bounds mismathed
anonymous
  • anonymous
Sorry that was actually suppose to be z=3y-x.
anonymous
  • anonymous
So the curve would equal C = xy^2 -3y+x
anonymous
  • anonymous
Sorry :(

Looking for something else?

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