Answer to Question #331311 in Calculus for Alfa

Question

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

Answer 1

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