I am just getting started with applications of integration, and I am stuck upon a footling question. Required to find the area bounded between $y^2=8x + 16$, $y^2=-4x + 28$, and the $x$-axis. The way I thought of solving this at first was to integrate with respect to $y$:
both curves have equal ordinates at points: $\smash[b]{\bigl( \frac{y^2+16}{8}, y \bigr)}$ for the curve $y^2=8x + 16$, and $\smash[b]{\bigl( \frac{28-y^2}{4}, y \bigr)}$ for the curve $y^2=-4x + 28$.
So the length of the horizontal strip connecting any two of those points is $$ \frac{28-y^2}{4} - \frac{y^2+16}{8} = 5 - \frac{3y^2}{8}. $$
The point of intersection of the two curves which is above the $x$-axis can be found to be $(1,\sqrt{24})$. So we are to calculate $$ \int_0^{\sqrt{24}} \biggl( 5 - \frac{3y^2}{8} \biggr) \, dy, $$ which is seen to be equal to $2\sqrt{24}$.
Is the procedure of my reasoning correct? Is the result acquired correct? I only consider myself a novice in integral calculus hence I seek constant affirmation from the more professional. Thank you in advance.
Your method is correct, but you made a small sign error in solving for $x$ for one of the boundary curves: $$ y^2 = 8x + 16 \iff x = \frac{y^2 \color{red}{-} 16}{8}, $$ hence the integrand is $$ \frac{28-y^2}{4} - \frac{y^2-16}{8} = \color{red}{9} - \frac{3y^2}{8}. $$ This produces the result $$ 6\sqrt{24} = 12\sqrt{6} \approx 29.394\ldots $$ which you can visualize here.
Incidentally, as a calculus instructor, I would reward you most of the credit for this answer, even though it's not correct, as you show clear understanding of the problem, a strategy for solving it, and even do most of the mechanics correctly too.