Answer:
[tex]\frac{16r^3}{3} [/tex]
Step-by-step explanation:
In this cross sections problem, we can integrate from -r to +r (so that the integral covers the entire base of the solid).
[tex]\int\limits^r_a {2(x^2+r^2 )} \, dx [/tex]
The formatting for the integral did not let me put -r on the lower bound, so i replaced it with a, just know that a represents -r here.
Evaluating the integral gives use that it is equal to;
[tex]\frac{16r^3}{3} [/tex]
Caution: this answer may not meet your needs, but this is the answer I have come up with with the given information.
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