Question #331311

Calculate the surface area Integral of object obtained by rotating the function x=2y+5 for x=-1 and -2 about y axis

1
Expert's answer
2022-04-20T14:52:50-0400

If the curve x=x(y)x=x(y) is rotated around yy-axis, the formula for the surface area is: S=2πy1y2x(y)1+(x(y))2dyS=2\pi\int_{y_1}^{y_2}x(y)\sqrt{1+(x'(y))^2}dy. In our case we have: x=2y+5,x=2y+5, y1=3,y_1=-3, y2=3.5y_2=-3.5. We get: S=2π33.55(2y+5)dy=2π5(y2+5y)y=3y=3.5=32π5S=2\pi\int_{-3}^{-3.5}\sqrt{5}(2y+5)dy=2\pi\sqrt{5}(y^2+5y)|_{y=-3}^{y=-3.5}=\frac32\pi\sqrt{5} . Thus, the answer is: S=32π5S=\frac32\pi\sqrt{5}.


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