# Let R be the region in the first quadrant bounded by the x-axis, the graph of x=y^2+2, and the line x=4. What is a the interval for the area of R?

Aug 28, 2017

$x$ goes from $2$ to $4$ and $y$ goes from $0$ to $\sqrt{2}$.

#### Explanation:

Sketch the graph of $x = {y}^{2} + 2$ by sketching the sideways parabola $x = {y}^{2}$ then adding $2$ to the $x$ coordinates (translate $2$ to the right).

Add the $x$ axis and the vertical line $x = 4$ to the picture:

The area can be found by either

${\int}_{2}^{4} \sqrt{x - 2} \textcolor{w h i t e}{\text{/}} \mathrm{dx}$

or by

${\int}_{0}^{\sqrt{2}} \left(4 - \left({y}^{2} + 2\right)\right) \textcolor{w h i t e}{\text{/}} \mathrm{dx}$