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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