Question

Question about an integral:

Answer 1

Why does:\[\int\limits_{0}^{1}(r+6r^2\sin(\theta)+8zr)dz = r(6rsin(\theta)+5)\]

I keep getting \(r(9+6rsin(\theta))\) as my answer but symbolab says its wrong?

Answer 2

Its part of a flux integral ive converted to polar

Answer 3

I distributed the r that was tacked on when I converted to polar, but I dont see how that would have affected it.

Answer 4

you can only seek help on this by posting the original question. The vector function etc :)

Answer 5

Sure.

Use the divergence theorem to find the total flux out of the solid x^2 + y^2 ≤ 16, 0 ≤ z ≤ 1 given v(x, y, z) = xi + 3y^2j + 4z^2k.

Answer 6

So you have \( \mathbf{v}(x, y, z) = x~ \hat i + 3y^2 ~ \hat j + 4z^2 \hat k\)


\(\nabla \cdot \mathbf v = 1 + 6 y +8 z\)

And then integrate inside that interval

Dunno.....that looks easy

Answer 7

Yea, thats right.

Answer 8

from there mate, it looks very do-able :)

Answer 9

FWIW, I did it this way for the practise and got \(80 \pi\). Seriously hope that's the "right answer" :)

\(\int_V 1 + 6y + 8z ~ dV\)

\(= \int\limits_{\theta = 0}^{2 \pi}~ \int\limits_{z = 0}^{1} ~ \int\limits_{r = 0}^{4} ~ ( 1 + 6r \sin \theta + 8z ) r dr~ dz ~ d \theta \)

\(= \int\limits_{\theta = 0}^{2 \pi}~ \int\limits_{z = 0}^{1} ~ \int\limits_{r = 0}^{4} ~ r + 6r^2 \sin \theta + 8zr ~ dr ~ dz~ d \theta\)

\(= \int\limits_{\theta = 0}^{2 \pi}~ \int\limits_{z = 0}^{1} \left[ r^2/2 + 2r^3 \sin \theta + 4zr^2 \right]_{r = 0}^{4} ~ dz ~ d \theta \)

\(= \int\limits_{\theta = 0}^{2 \pi}~ \int\limits_{z = 0}^{1} 8 + 128 \sin \theta + 64z ~ dz~ d \theta \)

\(= \int\limits_{\theta = 0}^{2 \pi} \left[ 8z + 128z \sin \theta + 32z^2 \right]_{z = 0}^{1} ~ dz\)

\(= \int\limits_{\theta = 0}^{2 \pi} 40 + 128 \sin \theta ~ d \theta \)

\(= \left[ 40 \theta - 128 \cos \theta \right]_{\theta = 0}^{2 \pi}\)

\(= 40 (2 \pi) - 128 \cos 2 \pi + 128 \cos 0 \)

\(= 80 \pi \)

the order of integration doesn't matter in this case, so maybe there is a easier order to compute.