Let’s go to new coordinates:
x˜=x;
y˜=y+1.
Now the volume of the solid can be obtained by rotating the region bounded by y˜=x˜2+2 and y˜=10−x˜2 about y˜=0.
Interval of integration:
x˜2+2=10−x˜2
x˜=±2
V=∫−22π((10−x˜2)2−(x˜2+2)2)dx˜=
∫−22π(96−24x˜2)dx˜=21∫02π(96−24x˜2)dx˜=
21π(96x˜−8x˜3)∣02=π(96−8⋅4)=64π
Answer: V=64π .
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