Question c62e7

May 8, 2016

The area is bounded by the maroon lines and the blue dots.
${A}_{T} = \frac{40}{3}$ Explanation:

The area is bounded by the maroon lines and the blue dots.
So we can break down the region into 2 parts: $\left[- 2 , 0\right) \cup \left[0 , 4\right]$
Note we are interested in the upper region so the area in each case is: $\text{ Area of rectangle - Area under the curve }$

$\left[{A}_{\square} - {A}_{y = {x}^{2}}\right] + \left[{A}_{\square} - {A}_{y = x}\right]$
$\left[{A}_{\square} - {A}_{y = {x}^{2}}\right] = 8 - {\int}_{-} {2}^{0} {x}^{2} \mathrm{dx} = \frac{16}{3}$
$\left[{A}_{\square} - {A}_{y = x}\right] = \frac{16}{2}$
${A}_{T} = \frac{16}{3} + \frac{16}{2} = \frac{40}{3}$

May 9, 2016

$\frac{40}{3}$

Explanation:

Referencing the picture in the other answer, we can split this into two integrals. The first is the region bounded on the interval $- 2 \le x \le 0$ by $y = 4$ and $y = {x}^{2}$. The area of this region can be found by integrating the upper function, $y = 4$, minus the lower function, $y = {x}^{2}$.

${\int}_{-} {2}^{0} \left(4 - {x}^{2}\right) \mathrm{dx} = {\left[4 x - {x}^{3} / 3\right]}_{-} {2}^{0}$

$= \left[4 \left(0\right) - {0}^{3} / 3\right] - \left[4 \left(- 2\right) - {\left(- 2\right)}^{3} / 3\right]$

$= 0 - \left[- 8 + \frac{8}{3}\right]$

$= \frac{16}{3}$

We have to add this value to the amount bounded by $y = x$ and $y = 4$, which is just a triangle.

The triangle has a height of $4$ and a length of $4$, which creates a triangle of area $8$.

The area of $8$ added to the other area of $\frac{16}{3}$ gives a total area of $\frac{40}{3}$.

May 10, 2016

Here is a third way to find the answer to this question. Rotate your thinking ${90}^{\circ}$

Explanation:

The region is shown below: We can think of the region as the collection of points whose $y$ coordinates vary from $0$ to $4$ while the $x$ coordinates go from a minimum of $- \sqrt{y}$ to a maximum of $y$. (The black horizontal lines in the image above.)

Note that the left branch of $y = {x}^{2}$ gives us $x = - \sqrt{y}$.

On the right it is easy to see that $y = x$ gives us $x = y$.

Our method is analogous to finding an area between $f \left(x\right)$ and $g \left(x\right)$ by integrating, with respect to $x$, the greater minus the lesser $y$ values,

we can integrate with respect to $y$:
Greater $x$ - lesser $x$ or
"$x$ on the right - $x$ on the left

${\int}_{0}^{4} \left(y - \left(- \sqrt{y}\right)\right) \mathrm{dy} = {\int}_{0}^{4} \left(y + \sqrt{y}\right) \mathrm{dy}$

 = y^2/2+2/3sqrty^3]_0^4

$= 8 + \frac{16}{3} = \frac{40}{3}$