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statuscvo [17]
3 years ago
7

Refer to the figure and find the volume V generated by rotating the given region about the specified line.

Mathematics
1 answer:
ANEK [815]3 years ago
7 0
Volume generated by the areas R2 + R3 around the line OC
= (1/3) pi 1^2 * 3  =  pi sq units

Volume generated by area R2 around OC  is found as follows

the  equation of the curve can be written as  x = y^4 / 81

Integral between limits y =  0 and 3  of  the curve is  

pi INT  x^2 dy    = INT pi  y^8 / 81^2  dx   

= pi [3^9 / 9*81^2] 

= pi/3

So the volume generated by R3 about OC = pi - pi/3  = 2pi/3 


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Anna007 [38]

Choice C is the correct answer because

\frac{(6+2)^3-12}{5}\\\\\frac{(8)^3-12}{5}\\\\\frac{512-12}{5}\\\\\frac{500}{5}\\\\100\\\\

So in short, \frac{(6+2)^3-12}{5}=100\\\\

------------------------------------------

The mistake Jerry likely made was that he only cubed the 2 and didn't realize the 6 was part of that cubing process. It seems he didn't add first and decided to cube before adding.

This is probably what steps Jerry did

\frac{6+2^3-12}{5}\\\\\frac{6+8-12}{5}\\\\\frac{14-12}{5}\\\\\frac{2}{5}\\\\

But as mentioned, those steps are incorrect because the 6 is part of the cubing operation. In other words, Jerry should have added the 6+2 first before cubing afterward (due to PEMDAS determining the order of operations).

Or you could think of it like this

\frac{(6+2)^3-12}{5}\\\\\frac{(6+2)(6+2)(6+2)-12}{5}\\\\\frac{(8)(8)(8)-12}{5}\\\\\frac{512-12}{5}\\\\\frac{500}{5}\\\\100

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Step-by-step explanation:

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