What is the area bounded by the curves? :  4x + y^2 = 32  and  x=y

Sep 23, 2017

We have:

$4 x + {y}^{2} = 32$
$x = y$

The graphs are as follows:

To find the coordinates of intersection:

$4 x + {x}^{2} = 32$
$\therefore \left(x - 4\right) \left(x + 8\right) = 0$
$\implies x = 4 , - 8$

So the intersection coordinates are:

$\left(- 8 , - 8\right)$ and $\left(4 , 4\right)$

We can calculate the bounded area (shaded) by integrating either wrt $x$ or wrt $y$, the latter being easier.

Method 1: Integrating wrt $y$

if we integrate with infinestimall thin horiozontal striops then we find the strips are bounded by $y = x$ on the left and by $4 x + {y}^{2} = 32$ on the right, and we integrate from the lower point of intersection$\left(- 8 , - 8\right)$ to the upper point of intersection $\left(4 , 4\right)$.

Thus, the area is given by:

$A = {\int}_{\alpha}^{\beta} \setminus f \left(y\right) \setminus \mathrm{dy}$
$\setminus \setminus \setminus = {\int}_{y = - 8}^{y = 4} \setminus \left(\frac{32 - {y}^{2}}{4}\right) - \left(y\right) \setminus \mathrm{dy}$
$\setminus \setminus \setminus = {\int}_{- 8}^{4} \setminus 8 - {y}^{2} / 4 - y \setminus \mathrm{dy}$
$\setminus \setminus \setminus = {\left[8 y - {y}^{3} / 12 - {y}^{2} / 2\right]}_{- 8}^{4}$
$\setminus \setminus \setminus = \left(32 - \frac{16}{3} - 8\right) - \left(- 64 + \frac{128}{3} - 32\right)$
$\setminus \setminus \setminus = \frac{56}{3} - \left(- \frac{160}{3}\right)$
$\setminus \setminus \setminus = 72$

Method 2: Integrating wrt $x$

if we integrate with infinestimall thin vertical striops then we find the strips are bounded by $4 x + {y}^{2} = 32$ at the bottom and $y = x$ at the top but only when $x$ is between the lower point of intersection$\left(- 8 , - 8\right)$ to the upper point of intersection $\left(4 , 4\right)$

We also need to include a portion that is bounded by $4 x + {y}^{2} = 32$ at the top and bottom between the upper point of intersection and the vertex (at $x = 8$).

Thus, the area is split into two seperate integrals:

$A = {A}_{1} + {A}_{2}$

Where:

${A}_{1} = {\int}_{x = - 8}^{x = 4} \left(x\right) - \left(- \sqrt{32 - 4 x}\right) \setminus \mathrm{dx}$
 A_2 =int_(x=4)^(x=8) \ (sqrt(32-4x) - (-sqrt(32-4x) ) \ dx

First we calculate ${A}_{1}$:

${A}_{1} = {\int}_{- 8}^{4} x + \sqrt{32 - 4 x} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus = {\left[{x}^{2} / 2 - \frac{4 {\left(8 - x\right)}^{\frac{3}{2}}}{3}\right]}_{- 8}^{4}$
$\setminus \setminus \setminus \setminus = \left(8 - \frac{32}{3}\right) - \left(32 - \frac{256}{3}\right)$
$\setminus \setminus \setminus \setminus = \left(- \frac{8}{3}\right) - \left(- \frac{160}{3}\right)$
$\setminus \setminus \setminus \setminus = \frac{152}{3}$

Then, ${A}_{2}$

${A}_{2} = {\int}_{4}^{8} \setminus 2 \sqrt{32 - 4 x} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus = 2 {\int}_{4}^{8} \setminus \sqrt{32 - 4 x} \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus = 2 {\left[- \frac{4 {\left(8 - x\right)}^{\frac{3}{2}}}{3}\right]}_{4}^{8}$
$\setminus \setminus \setminus \setminus = 2 \left\{\left(0\right) - \left(- \frac{32}{3}\right)\right\}$
$\setminus \setminus \setminus \setminus = \frac{64}{3}$

So, the total area is:

$A = {A}_{1} + {A}_{2}$
$\setminus \setminus \setminus = \frac{152}{3} + \frac{64}{3}$
$\setminus \setminus \setminus = 72$