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maria [59]
3 years ago
10

Can you help me with this please

Mathematics
1 answer:
djyliett [7]3 years ago
5 0
The answer to this problem is simple all you have to do is pemdas and you get the answer 0.6
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Read 2 more answers
Integrate f(x,y,z)=12xz over the region in the first octant above the parabolic cylinder z=y^2 and below the paraboloid z=8−2x^2
Neko [114]
The intersection of the two surfaces occurs along the cylinder x^2+y^2=4, so we have the x- and y-coordinates of the points in the bounded region contained within 0\le x\le2 and 0\le y\le2, while y^2\le z\le8-2x^2-y^2.

The triple integral is then given by

\displaystyle\iiint_V12xz\,\mathrm dV=\int_{x=0}^{x=2}\int_{y=0}^{y=2}\int_{z=y^2}^{z=8-2x^2-y^2}12xz\,\mathrm dz\,\mathrm dy\,\mathrm dx

where V denotes the bounded region in the first quadrant.

Integrating in Cartesian coordinates is easy enough to do in this order.

\displaystyle\int_{x=0}^{x=2}\int_{y=0}^{y=2}\int_{z=y^2}^{z=8-2x^2-y^2}12xz\,\mathrm dz\,\mathrm dy\,\mathrm dx
=\displaystyle24\int_{x=0}^{x=2}\int_{y=0}^{y=2}x(x^2-4)(x^2+y^2-4)\,\mathrm dy\,\mathrm dx
=\displaystyle\int_{x=0}^{x=2}(512x-320x^3+48x^5)\,\mathrm dx
=256
3 0
3 years ago
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