Described area:
Then
V = π ∫ a b ( f 2 ( x ) − g 2 ( x ) ) d x = π ∫ − 2 3 ( ( x + 8 ) 2 − ( x 2 + 2 ) 2 ) d x = V = \pi \int\limits_a^b {\left( {{f^2}(x) - {g^2}(x)} \right)dx} = \pi \int\limits_{ - 2}^3 {({{(x + 8)}^2} - {{({x^2} + 2)}^2}} )dx = V = π a ∫ b ( f 2 ( x ) − g 2 ( x ) ) d x = π − 2 ∫ 3 ( ( x + 8 ) 2 − ( x 2 + 2 ) 2 ) d x =
= π ∫ − 2 3 ( x 2 + 16 x + 64 − x 4 − 4 x 2 − 4 ) d x = π ∫ − 2 3 ( − x 4 − 3 x 2 + 16 x + 60 ) d x = = \pi \int\limits_{ - 2}^3 {\left( {{x^2} + 16x + 64 - {x^4} - 4{x^2} - 4} \right)dx = } \pi \int\limits_{ - 2}^3 {\left( { - {x^4} - 3{x^2} + 16x + 60} \right)dx = } = π − 2 ∫ 3 ( x 2 + 16 x + 64 − x 4 − 4 x 2 − 4 ) d x = π − 2 ∫ 3 ( − x 4 − 3 x 2 + 16 x + 60 ) d x =
= π ( − x 5 5 ∣ − 2 3 − x 3 ∣ − 2 3 + 8 x 2 ∣ − 2 3 + 60 x ∣ − 2 3 ) = = \pi \left( { - \left. {\frac{{{x^5}}}{5}} \right|_{ - 2}^3 - \left. {{x^3}} \right|_{ - 2}^3 + \left. {8{x^2}} \right|_{ - 2}^3 + 60\left. x \right|_{ - 2}^3} \right) = = π ( − 5 x 5 ∣ ∣ − 2 3 − x 3 ∣ ∣ − 2 3 + 8 x 2 ∣ ∣ − 2 3 + 60 x ∣ − 2 3 ) =
= π ( − 243 + 32 5 − ( 27 + 8 ) + 8 ( 9 − 4 ) + 60 ( 3 + 2 ) ) = π ( − 55 − 35 + 40 + 300 ) = 250 π = \pi \left( { - \frac{{243 + 32}}{5} - (27 + 8) + 8(9 - 4) + 60(3 + 2)} \right) = \pi \left( { - 55 - 35 + 40 + 300} \right) = 250\pi = π ( − 5 243 + 32 − ( 27 + 8 ) + 8 ( 9 − 4 ) + 60 ( 3 + 2 ) ) = π ( − 55 − 35 + 40 + 300 ) = 250 π
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