The area of the region enclosed by the curves y=x^2-4 x+4 and y^2=16-8 x is :

Question

The area of the region enclosed by the curves $y=x^2-4 x+4$ and $y^2=16-8 x$ is :

Accepted Answer

Consider the curves

$y = x^2 - 4x + 4 = (x-2)^2$

and

$y^2 = 16 - 8x.$

Notice that the second equation can be rewritten in terms of $x$:

$$ 8x = 16 - y^2 \quad \Longrightarrow \quad x = 2 - \frac{y^2}{8}. $$

Find the Intersection Point: 1. Find the Intersection Points

To find the points where the curves intersect, substitute

$y = (x-2)^2$

into

$y^2 = 16 - 8x.$

Let

$$ u = x - 2 \quad \text{so that} \quad y = u^2. $$

Then

$y^2 = u^4$

and

$x = u + 2.$

Substitute into the second curve:

$u^4 = 16 - 8(u+2).$

Simplify the right side:

$16 - 8(u+2) = 16 - 8u - 16 = -8u.$

Thus, the equation becomes

$$ u^4 + 8u = 0 \quad \Longrightarrow \quad u(u^4/ u? \,\, {correcting factor}). $$

In fact, factor by taking out a common factor $u$:

$u(u^3 + 8) = 0.$

Thus, either

$$ u = 0 \quad \text{or} \quad u^3 = -8. $$

For $u = 0$:

$$ x = u + 2 = 2, \quad y = u^2 = 0. $$

For $u^3 = -8$:

$$ u = -2, \quad \text{so} \quad x = -2 + 2 = 0, \quad y = (-2)^2 = 4. $$

The curves intersect at the points $(2, 0)$ and $(0, 4).$

Express the Curves in Terms of $$y: 2. Express the Curves in Terms of $$y$$

It is easier to integrate horizontally by expressing $$x$as a function of$y$.

From the second curve:$ x = 2 - \frac{y\^2}{8}. $From the first curve, solving$ y = (x-2)\^2 $for$x$gives$ x-2 = \pm\sqrt{y}. $Since at the intersection$ (0, 4) $the$x$$--value is less than 2, we take the negative branch:

$$ x = 2 - \sqrt{y}. $Thus, for a fixed$y$between 0 and 4, the left boundary is$ x\_{\text{left}} = 2 - \sqrt{y}, $and the right boundary is$ x\_{\text{right}} = 2 - \frac{y\^2}{8}. $$

Set Up the Integral for the Are: 3. Set Up the Integral for the Area

The horizontal distance between the curves at a given $$y$is$ \Delta x = x\_{\text{right}} - x\_{\text{left}} = \left(2 - \frac{y\^2}{8}\right) - \left(2 - \sqrt{y}\right) = \sqrt{y} - \frac{y\^2}{8}. $Integrate with respect to$y$from$y = 0$to$y = 4$:$ A = \int\_{0}\^{4} \left(\sqrt{y} - \frac{y\^2}{8}\right) \, dy. $$

Evaluate the Integra: 4. Evaluate the Integral

Write the integral as

$$ A = \int\_{0}\^{4} y\^{1/2}\, dy - \frac{1}{8} \int\_{0}\^{4} y\^2\, dy. $$

Compute each term:

For the first integral:

$$ \int y\^{1/2}\, dy = \frac{2}{3} y\^{3/2}. $For the second integral:$ \int y\^2\, dy = \frac{y\^3}{3}. $Thus,$ A = \left[\frac{2}{3} y\^{3/2}\right]\_0\^4 - \frac{1}{8}\left[\frac{y\^3}{3}\right]\_0\^4. $Substitute$y = 4$(note that at$y = 0$both terms vanish):

Compute$4\^{3/2}$:$ 4\^{3/2} = \left( \sqrt{4} \right)\^3 = 2\^3 = 8. $Compute the first term:$ \frac{2}{3} \times 8 = \frac{16}{3}. $Compute the second term:$ \frac{1}{8} \cdot \frac{64}{3} = \frac{64}{24} = \frac{8}{3}. $Therefore, the area is$ A = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}. $$

Final Answer

The area of the region enclosed by the curves is

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