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3 years ago

Can someone please help me

Mathematics

1 answer:

Vsevolod [243]3 years ago

5 0

The restrictions for the equation is that the denominator can not be zero. So the restrictions for x would be what values make the denominator zero.

7x^2 + 6x = 0
factor
x(7x + 6) = 0
multiply any number by 0 and you get 0. So either x = 0 or 7x+6 = 0
since there's a x in the numerator x/x = 1 so this will not be a restriction. Then the only restriction is:
7x+6 = 0
7x = -6
x = -6/7

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Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant

Alex787 [66]

Find the intercepts for both planes.

Plane 1, x + y + 2z = 2:

$y=z=0\implies x=2\implies (2,0,0)$

$x=z=0\implies y=2\implies(0,2,0)$

$x=y=0\implies 2z=2\implies z=1\implies(0,0,1)$

Plane 2, 4x + 4y + z = 8:

$y=z=0\implies4x=8\implies x=2\implies(2,0,0)$

$x=z=0\implies4y=8\impliesy=2\implies(0,2,0)$

$x=y=0\implies z=8\implies(0,0,8)$

Both planes share the same x- and y-intercepts, but the second plane's z-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (x, y)-plane where z = 0, we see the bounded region projects down to the triangle in the first quadrant with legs x = 0, y = 0, and x + y = 2, or y = 2 - x.

So the volume of the region is

$\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx$

$=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}$

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