3 years ago

99 POINT QUESTION PLUS BRAINLIEST!!!

1 answer:

7 0

The graphs arey=(1/3)x^2
and y=3
they intersect at x=3
so the limits are x=0 and 3
to rotate it about y=k, minus k from every function

$area=\pi \int\limits^3_0 {( \frac{1}{3}x^2-3)^2-(3-3)^2 } \, dx$
which evaluates to 72/5pi

A is asnwer

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