Question #140560
The area is bounded by the curve y=x^2+2 and the line y=x+8 is rotated around the x axis. Find the volume of the solid generated.
1
Expert's answer
2020-10-27T18:35:53-0400

Described area:



Then

V=πab(f2(x)g2(x))dx=π23((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 =

=π23(x2+16x+64x44x24)dx=π23(x43x2+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(x5523x323+8x223+60x23)== \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(243+325(27+8)+8(94)+60(3+2))=π(5535+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


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