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3 years ago

6

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

e y-axis.
y = x^2, y = 10 - 9x, x = 0, for x ≥ 0

1 answer:

I am Lyosha [343]3 years ago

7 0

Answer:

<h2>it is 6 because it's hard to explain but all I know it's 6</h2>

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3 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$

A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095

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2 years ago