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egoroff_w [7]
3 years ago
15

I need serious helpplease its confusing....

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

area: 40p² + 24p

wide: 8p

(40p² + 24p)/ 8p

5p + 3

so, length: 5p + 3 -> (a)

perimeter = wide x 2 + length x 2

16p + (10p + 6)

26p + 6 -> (b)

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Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about th
RSB [31]

Answer:

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\displaystyle\frac{19\pi}{6}

Step-by-step explanation:

See the graph of the region attached.

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x^2=9-8x\\x^2+8x-9=0\\(x+9)(x-1)=0\\x=-9,x=1

Since the region is the one for x\ge 0 then the intersection we are interested on is x=1 as it can also be seen in the graph.

Then we set the integral using shell method for revolving about the y-axis:

\displaystyle\int_a^b 2\pi\,x\,h(x)\,dx

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Then the integral becomes:

\displaystyle\int_0^1 2\pi\,x\,(8x-9-x^2)\,dx

Notice the limits of the integral are the x-axis (x=0) and the intersection of the parabola and the line that we found before (x=1)

Now, solving the integral:

Start by factoring the 2\pi and distributing the x:

=\displaystyle2\pi \int_0^19x-8x^2-x^3dx

Then use the basic rule to integrate:

=\displaystyle2\pi \left[\frac{9x^2}{2}-\frac{8x^2}{3}-\frac{x^4}{4}\right|_0^1

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4 0
3 years ago
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