Calculate the volume under the elliptic paraboloid z = 2x^2 + 4y^2 and over the rectangle R = [−3, 3] ×[−1, 1].

if i see it right, everything seems to be centered about the origin

ok !

if we calculate the volume in the first "quadrant" and quadruple it; that should get us all of it

\[\int_{0}^{3}\int_{0}^{1}2x^2+4x^2dydx\] is what im thinking

its either that or i got my xs and ys backwards

hmm..ok

does this look familiar? or am i way off base from what your studying?

how do i use the fundamental of calculus to calculate the numerical value !?

im not sure how the fundamental of calculus would word into this, i just tend to apply it an dhope for the best.

well i mean how do u solve that double integral then

the first step, if im reading it right would be to work thru the inner integral: \[\int_{0}^{3}\int_{0}^{1}2x^2+4x^2dydx\] \[\int_{0}^{1}2x^2+4x^2dy=2x^2y+\frac{4}{3}y^3;[0,1]\] \[2x^2(1)+\frac{4}{3}(1)^3-2x^2(0)-\frac{4}{3}(0)^3\] \[2x^2+\frac{4}{3}\] then work on the outer integral with these results \[\int_{0}^{3}2x^2+\frac{4}{3}dx\]

i appear to have typoed over the y part and implanted it back in ... how odd of me

ok let me calculate this ! and confirm it with u

22 ?

22 looks good, and i come up with that when i swap my xs and ys as well; so this is 1/4 the volume as i see it; so times it by 4

i pluged them both in and both r wrong !

\[\int_{0}^{1}\int_{0}^{3}2x^2+4x^2dydx\] then i might try it in this manner; and if that fails then I just aint reading it right

4x^2 means 4y^2 of course

so should i do the same thing with y and the integral from 0 to 1 ?

thats what we did it the start and came up wrong i believe; i figure this one to be 152

the caveat im having is the R, im assuming that is a rectangled bound on the xy plane

so its 152 !? or 152/4 ?!

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