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TEA [102]
3 years ago
8

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

out the x-axis. (Round your answer to three decimal places.)
Mathematics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

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Answer:

Mean = 75

Median = 73.5

Mode = 95

Range = 36

Step-by-step explanation:

Given:

  • 77,60,59,70,89,95

Sort:

  • 59,60,70,77,89,95

To find:

  • Mean
  • Median
  • Mode
  • Range

Mean:

\displaystyle \large{\dfrac{1}{n}\sum_{i =1}^n x_i = \dfrac{x_1+x_2+x_3+...+x_n}{n}}

Sum of all data divides by amount.

\displaystyle \large{\dfrac{59+60+70+77+89+95}{6}=\dfrac{450}{6}}\\\\\displaystyle \large{\therefore mean=75}

Therefore, mean = 75

Median:

If it’s exact middle then that’s the median. However, if two data or values happen to be in <em>middle</em>:

\displaystyle \large{\dfrac{x_1+x_2}{2}}

From 59,60,70,77,89,95, since 70 and 77 are in middle:

\displaystyle \large{\dfrac{70+77}{2} = \dfrac{147}{2}}\\\displaystyle \large{\therefore median = 73.5}

Therefore, median = 73.5

Mode:

The highest value or/and the highest amount of data. Mode can have more than one.

From sorted data, there are no repetitive data nor same data. Consider the highest value:

Therefore, mode = 95

Range:

\displaystyle \large{x_{max}-x_{min}} or highest value - lowest value

Thus:

\displaystyle \large{95-59 = 36}

Therefore, range = 36

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