Surface Area — Question 3

Question 3

Curve: $y=\sqrt{4-x^2}$ on $0\le x\le2$.
Axis of rotation: the $x$-axis.
Task: Identify the resulting surface and find its area.

Question 3 – Solution

Step 1: Identify the surface. The curve is the upper-right quarter of the circle $x^2+y^2=4$. Rotating it about the $x$-axis produces a hemisphere of radius (2).

Step 2: Differentiate. $$y'=-\frac{x}{\sqrt{4-x^2}}.$$ Step 3: Simplify the arc-length factor. $$\begin{align*} \sqrt{1+(y')^2} &=\sqrt{1+\frac{x^2}{4-x^2}}\\ &=\sqrt{\frac4{4-x^2}}=\frac2{\sqrt{4-x^2}}. \end{align*}$$ Step 4: Calculate the surface area. $$\begin{align*} S&=2\pi\int_0^2y\sqrt{1+(y')^2}\,dx\\ &=2\pi\int_0^2\sqrt{4-x^2}\frac2{\sqrt{4-x^2}}\,dx\\ &=4\pi\int_0^2dx=8\pi. \end{align*}$$ $$\boxed{\text{hemisphere},\ S=8\pi}$$