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nignag [31]
3 years ago
9

Find area and circumference of the circle (12.) pls NO links

Mathematics
2 answers:
Irina-Kira [14]3 years ago
7 0
Area - 452.39
Circumference - 75.4
xxTIMURxx [149]3 years ago
5 0

Answer:

The area is 452.39

The Circumference is 75.36

Step-by-step explanation:

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Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the
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Answer:

Required volume is \frac{4864\pi}{3} unit.

Step-by-step explanation:

Given equations of curves,

y=32-x^2, y=x^2

substitute second in first we will get,

x^2=36-x^2\implies x^2=16\implies x=\pm 4

When x=4, y=16 and x=-4, y=16. Thus both curves intersect at the points (4,16),(-4,16). And thus -4.

Considering width of the representative rectangle as \Delta x which is parallel to x-axis because of considering cylindrical shell. Therefore volume of the shell is given by,

V_{Shell}=(\textit{Length of box})\times (\textit{Width of box})\times (\textit{Thikness of box})

Now,

Length of generating box=Length of \Delta x from line x=4(axis of revolution)=circumference of the shell=(4-x)

Width of box=(36-x^2)-x^2

Hence,

V_{Shell}

=\int_{-4}^{4}2\pi (4-x)(36-2x^2)dx

=2\pi\int_{-4}^{4}(144-36x-8x^2+2x^3)dx

=2\pi\{144\big[x\big]_{-4}^{4}-18\big[x^2\big]_{-4}^{4}-\frac{8}{3}\big[x^3\big]_{-4}^{4}+\frac{1}{2}\big[x^4\big]_{-4}^{4}\}

=\frac{4864\pi}{3}

which is required volume of generating area.

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