The discussion centers on calculating the surface area of a solid of revolution defined by the equation \(x = \frac{1}{3} (y^2 + 2)^{3/2}\) for \(1 \leq y \leq 3\) about the x-axis. The user correctly applies the formula for surface area, integrating in terms of \(y\) and using the derivative \(dx/dy\) to find the necessary components for the integral. The final result of the surface area calculation is confirmed to be \(48\pi\), validating the user's approach and calculations.
PREREQUISITESStudents and professionals in mathematics, particularly those studying calculus and its applications in geometry, as well as educators teaching surface area concepts in advanced mathematics courses.
ineedhelpnow said:that's how i did the problem:
$x=\frac{1}{3} (y^2+2)^{3/2}$ $1\le y \le 3$ about the x axis
$x'= \frac{(y^2+2)^{3/2}}{3} = \frac{ \frac{3(y^2+2)^{1/2}}{2}}{3}dy(y^2+2 ) = \frac{(y^2+2)^{1/2}*2y}{2} = y \sqrt{y^2+2}$
$S=\int_{1}^{3} \ 2\pi y \sqrt{1+y^2(y^2+2)}dy=2\pi \int_{1}^{3} \ y\sqrt{1+y^4+2y^2}dy=2\pi\int_{1}^{3} \ y\sqrt{(y^2+1)^2}dy=2\pi \int_{1}^{3} \ y(y^2+1)dy=2\pi \int_{1}^{3} \ (y^3+y) dy=2\pi [\frac{y^4}{4}+\frac{y^2}{2}]_{1}^{3} = 48 \pi$
is it correct?