Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the given axis. y-axis

Question

Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the given axis.  y-axis

Quizplus · Accepted Answer

We need to use the formula for the surface area obtained by rotating the curve $y=f(x)$ about the $y$-axis, which is $S = 2\pi \int_a^b f(x) \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$. From the given figure, we see that the curve is given by $y=\frac{x^2}{2}$ and $x=y^2-1$. We need to find the limits of integration $a$ and $b$. Solving for $x$ in terms of $y$ in the equation $x=y^2-1$, we get $x+1=y^2$, which gives $y=\pm \sqrt{x+1}$, so the curve is traced out from $y=-\sqrt{x+1}$ to $y=\sqrt{x+1}$. Thus, our integral is $$ \int_{-1}^{0} 2\pi \left(\frac{x^2}{2}\right) \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx + \int_{0}^{2} 2\pi \left(\frac{x^2}{2}\right) \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx.$$Next, we need to find $\frac{dy}{dx}$. Differentiating the equation $y=\frac{x^2}{2}$ with respect to $x$, we get $\frac{dy}{dx} = x$. Substituting this into $\frac{dy}{dx}$ in the equation $x=y^2-1$, we get $\frac{dy}{dx} = 2y$. Thus, we have $$S = 2\pi\left(\int_{-1}^{0} \frac{x^2}{2} \sqrt{1+x^2} dx + \int_{0}^{2} \frac{x^2}{2} \sqrt{1+4x^2} dx\right).$$This matches option (B), so the answer is (B).

Set Up, but Do Not Evaluate, an Integral for the Area