this should be easy....... Find the volume of the solid enclosed by the paraboloids z=1(x^2+y^2) and z=18−1(x^2+y^2).
set the two equal and solve for r gets r=3. so i have...\[\int\limits_{0}^{2\pi}\int\limits_{0}^{3}18-2r^2 r dr dtheta\]giving an answer of 99pi/4. webwork says no. i just did this in the double integral section.....
hate to be a bother, but @ganeshie8, could you help me again?
that looks good to me, except for a missing parenthesis
\[\int\limits_{0}^{2\pi}\int\limits_{0}^{3}(18-2r^2) r dr dtheta\]
yeah, i put that into the wolfram alpha widget for polar.... http://www.wolframalpha.com/widgets/view.jsp?id=819a2d24e73f94fa5a05de2fad9ebddc
webwork says that answer is incorrect
well, i tried using that two ways. the firs time was 0 to 2pi, but it gave the same answer.
im getting 81pi ?
oh. i didn't know you had to put () around the equation in the first box or the widget would do it wrong.....grr. thanks.
yeah the widgest is simply attaching rdrdtheta at the end it seems
which is weird
it should multiply r by the entire integrand, not just the last term
i'll know that for next time. now i'm battling writing a triple integral six different ways.....
3! = 6
does your professor really hates u that much ?
lol. i guess. Express the integral ∭Ef(x,y,z)dV as an iterated integral in six different ways, where E is the solid bounded by z=0,x=0,z=y−x and y=2. i got the first four. struggling with dydzdx and dydxdz
|dw:1415018561570:dw|
should be 0 to 2, 0 to x, z+x to 2. but it doesn't like the 0 to x part.....can't figure out why.
for dydzdx : x : 0->2 z : 0->-x y : z+x->2 for dydxdz : z : 0->2 x : 0->-z y : z+x->2
see if they work
the -x and -z answers didn't work, @ganeshie8 . sorry for the delay, i just got home from work.
could u take a screenshot and attach again
yes, beccaboo. this is for the problem: Express the integral ∭Ef(x,y,z)dV as an iterated integral in six different ways, where E is the solid bounded by z=0,x=0,z=y−x and y=2.
try entering -x+2
and -z+2
respectively for the incorrect ones
ugh. finally! thank you, becca!
you're welcome :)
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