The volume generated by rotating the given region [tex]R_{3}[/tex] about OC is [tex]\frac{4}{g} \pi[/tex]
Washer method
Because the given region ([tex]R_{3}[/tex]) 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
[tex]y=2(x)^{\frac{1}{4} }[/tex]
[tex]x=(\frac{y}{2} )^{4}[/tex]
[tex]y=2x[/tex]
[tex]x=\frac{y}{2}[/tex]
[tex]\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy[/tex]
[tex]R_{o} = x = \frac{y}{2}[/tex]
[tex]R_{i} = x= (\frac{y}{2}) ^{4}[/tex]
[tex]a=0, b=2[/tex]
[tex]v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy[/tex]
[tex]v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy[/tex]
[tex]v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ][/tex]
[tex]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[/tex]
A similar question about finding the volume generated by a given region is answered here: https://brainly.com/question/3455095
