Area of a surface by rotating the curve about the x-axis. The curve $y=\sqrt{5-x}$ with $a=3$ and $b=5$ is rotated about the $x$-axis. Find the exact area of the surface obtained. 

In order to solve this problem, we need to use the following equation: $$ SA = 2\pi \int_{a}^{b}y\sqrt{1+(\frac{dy}{dx})^2}\hspace{1mm}dx $$ Where y, in this case, is given by: $$ y = \sqrt{5-x} $$ And, as you mentioned in your comment, the derivative with respect to x is given by: $$\frac{dy}{dx}=\frac{-1}{2\sqrt{5-x}} $$We then can substitute these expressions into the equation: $$ SA = 2\pi \int_{3}^{5}\sqrt{5-x}\hspace{0.6mm}\cdot \hspace{0.6mm} \sqrt{1+\frac{1}{4(5-x)}} \hspace{1mm}dx $$ Since the left term and the right term both have an exponent of ½, and we know that: $$ a^c \cdot b^c=(ab)^c  $$ We can simplfy the integrand to: $$ \sqrt{5-x+\frac{1}{4}}  $$ Which leaves us with the following expression to be integrated:$$ 2\pi \int_{3}^{5} \sqrt{\frac{21}{4}-x} \hspace{1mm}dx $$ Which shouldn't be too difficult of an integral to solve at this point in your calculus studies. 
