Surface Area — Question 4

Question 4 – Solution

Step 1: Find the radius and derivative. Rotation about the $y$-axis gives radius $r=x$, and $$y'=\frac1{2\sqrt{x}}.$$ Step 2: Set up and simplify. $$\begin{align*} S&=2\pi\int_1^4x\sqrt{1+\frac1{4x}}\,dx\\ &=\pi\int_1^4\sqrt{x(4x+1)}\,dx. \end{align*}$$ Step 3: Substitute $u=8x+1$. Then $$dx=\frac{du}{8},\qquad\sqrt{x(4x+1)}=\frac14\sqrt{u^2-1}.$$ Thus $$S=\frac\pi{32}\int_9^{33}\sqrt{u^2-1}\,du.$$ Step 4: Integrate and return to $x$. $$\int\sqrt{u^2-1}\,du =\frac12\left(u\sqrt{u^2-1}-\ln(u+\sqrt{u^2-1})\right).$$ Therefore, $$\boxed{S=\pi\left[ \frac{(8x+1)\sqrt{x(4x+1)}}{16} -\frac1{64}\ln\left(8x+1+4\sqrt{x(4x+1)}\right) \right]_1^4}$$