Let $R$ be the region enclosed by the curves $x = y^2 + 1, x = 5, y = −3$ and $y = 3$

Question

(i) Sketch the region $R$. (ii) Find the area of region $R$.

Can someone tell me if i am doing it correctly since i tried it out and sketch a graph like the one on the picture but the points $y=3$ and $y=-3$ makes me confuse since i did my sketch but never use those two points also the question never mentioned if the function revolve about either $x$-axis or $y$-axis.

Answer 1

There are two curves $$y=\sqrt {x-1} $$ and $$y=-\sqrt {x-1} $$ $$y=\pm 3\implies x=10$$ By symetry, the area between them is $$2\int_5^{10}\sqrt {x-1}dx=$$ $$=2 \Bigl[\frac {2}{3}(x-1)^\frac 32\Bigr]_5^{10} $$ $$\frac {4}{3}(27-8)=\frac {76}{3} $$

Using Donald's graph, the area you look for is

$$(10-5)(3+3)-\frac {76}{3}=\frac {14}{3} $$

Answer 2

This would be my sketch:

And I would calculate the area to be:

$\displaystyle {\int_{-3}^{-2} (y^2 +1) - 5 \ dy + \int_{-2}^{2} 5-(y^2 +1)\ dy + \int_{2}^{3} (y^2 +1)-5\ dy}$