Described area:
Then
V=π∫ab(f2(x)−g2(x))dx=π∫−23((x+8)2−(x2+2)2)dx=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=πa∫b(f2(x)−g2(x))dx=π−2∫3((x+8)2−(x2+2)2)dx=
=π∫−23(x2+16x+64−x4−4x2−4)dx=π∫−23(−x4−3x2+16x+60)dx== \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(x2+16x+64−x4−4x2−4)dx=π−2∫3(−x4−3x2+16x+60)dx=
=π(−x55∣−23−x3∣−23+8x2∣−23+60x∣−23)== \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(−5x5∣∣−23−x3∣∣−23+8x2∣∣−23+60x∣−23)=
=π(−243+325−(27+8)+8(9−4)+60(3+2))=π(−55−35+40+300)=250π= \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(−5243+32−(27+8)+8(9−4)+60(3+2))=π(−55−35+40+300)=250π
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