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Natasha2012 [34]
3 years ago
10

Express the integral as an iterated integral in six different ways, where E is the solid bounded by y=4-x^2-4z^2 and y=0

Mathematics
1 answer:
zmey [24]3 years ago
5 0
Assuming you need the integral expressing the volume of E, the easiest setup is to integrate with respect to y first.

This is done with either

\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dx\,\mathrm dz
\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dz\,\mathrm dx

Thanks to symmetry, integrating with respect to either x or z first will be nearly identical.

First, with respect to x:

\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dy\,\mathrm dz
\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dz\,\mathrm dy

Next, with respec to z:

\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dy\,\mathrm dx
\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dx\,\mathrm dy
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Serhud [2]

Answer:

$3

Step-by-step explanation:

Given data

Let the cost of drinks be x

and the cost of hotdogs be y

so

3x+4y=15-------1

also

3x+y= 10.5------2

solve 1 and 2

 3x+4y=15

-3x+y= 10.5

     3y= 4.5

divide both sides by 3

y= 4.5/3

y= $1.5

Put y= 1.5 in 1

3x+4(1.5)=15

3x+6= 15

3x=15-6

3x= 9

x= 9/3

x= $3

Hence the cost of a drink is $3

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3 years ago
What is the volume of the prism?
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For  the volume, the formula is length*width*height
so
<span><span><span>(2)</span><span>(1.5)</span></span><span>(1.5)</span></span><span>=<span>4.5
In other words, your answer would be 4.5</span></span>
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3 years ago
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Simplify the following surd expressions<br> a) 7/3 - 2/3 + V3 - 3V3
Nata [24]

Answer:

\frac{7}{3}-\frac{2}{3}+\sqrt{3}-3\sqrt{3}=\frac{5}{3}-2\sqrt{3}    

Step-by-step explanation:

Given the expression

\frac{7}{3}\:-\:\frac{2}{3}\:+\:v^3\:-\:3v^3

solving the expression

\frac{7}{3}\:-\:\frac{2}{3}\:+\:\sqrt{3}-3\sqrt{3}

combine the fractions i.e \frac{7}{3}-\frac{2}{3}=\frac{5}{3}

\frac{7}{3}\:-\:\frac{2}{3}\:+\:\sqrt{3}-3\sqrt{3}=\frac{5}{3}+\sqrt{3}-3\sqrt{3}

add similar elements  i.e \sqrt{3}-3\sqrt{3}=-2\sqrt{3}

                                 =\frac{5}{3}-2\sqrt{3}      

Thus,

\frac{7}{3}-\frac{2}{3}+\sqrt{3}-3\sqrt{3}=\frac{5}{3}-2\sqrt{3}                        

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