Blog: `4x + y^2 = 12, x = y` Sketch the region enclosed by the given curves. Decide whether to integrate with respect to `x` or `y`. Draw a typical...

Wednesday, August 31, 2016

$\displaystyle{4}{x}+{{y}}^{{2}}={12},{x}={y}$ Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $\displaystyle{x}$ or $\displaystyle{y}$ . Draw a typical...

You need to determine first the points of intersection between curves $\displaystyle{y}=\sqrt{{{12}-{4}{x}}}$ and $\displaystyle{y}={x}$ , by solving the equation, such that:

$\displaystyle\sqrt{{{12}-{4}{x}}}={x}\Rightarrow{12}-{4}{x}={{x}}^{{2}}\Rightarrow{{x}}^{{2}}+{4}{x}-{12}={0}$

$\displaystyle{x}_{{{1},{2}}}=\frac{{-{4}\pm\sqrt{{{16}+{48}}}}}{{2}}\Rightarrow{x}_{{{1},{2}}}=\frac{{-{4}\pm{8}}}{{2}}$

$\displaystyle{x}_{{1}}={2};{x}_{{2}}=-{6}$

Hence, the endpoints of integral are x = -6 and x = 2

You need to decide what curve is greater than the other on the interval [-6,2]. You need to notice that $\displaystyle\sqrt{{{12}-{4}{x}}}>{x}$ on the interval [-6,2],hence, you may evaluate the area of the region enclosed by the given curves, such that:

$\displaystyle{\int_{{a}}^{{b}}}{\left({f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right)}{\left.{d}{x}\right.}$ , where f(x) > g(x) for $\displaystyle{x}\in{\left[{a},{b}\right]}$

$\displaystyle{\int_{{-{6}}}^{{2}}}{\left(\sqrt{{{12}-{4}{x}}}-{x}\right)}{\left.{d}{x}\right.}={\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}-{\int_{{-{6}}}^{{2}}}{x}{\left.{d}{x}\right.}$

You need to solve $\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}$ using substitution $\displaystyle{12}-{4}{x}={t}$ , such that:

$\displaystyle{12}-{4}{x}={t}\Rightarrow-{4}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{\left.{d}{x}\right.}=-\frac{{{\left.{d}{t}\right.}}}{{4}}$

$\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=-{\left(\frac{{1}}{{4}}\right)}{\int_{{{t}_{{1}}}}^{{{t}_{{2}}}}}\sqrt{{{t}}}{\left.{d}{t}\right.}$

$\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=-\frac{{2}}{{12}}{{\left({12}-{4}{x}\right)}}^{{\frac{{3}}{{2}}}}{{\mid}_{{-{6}}}^{{2}}}$

$\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=-\frac{{2}}{{12}}{{\left({12}-{4}\cdot{2}\right)}}^{{\frac{{3}}{{2}}}}+\frac{{2}}{{12}}{\left({12}+{24}\right)}$

$\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=-{\left(\frac{{2}}{{12}}\right)}\cdot{8}+\frac{{{36}\cdot{2}}}{{12}}$

$\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=-\frac{{4}}{{3}}+{6}$

$\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=\frac{{{18}-{4}}}{{3}}$

$\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=\frac{{14}}{{3}}$

Hence, evaluating the area of the region enclosed by the given curves, yields $\displaystyle{\int_{{-{6}}}^{{2}}}\sqrt{{{12}-{4}{x}}}{\left.{d}{x}\right.}=\frac{{14}}{{3}}.$

The area of the region enclosed by the given curves is found between the red and orange curves, for $\displaystyle{x}\in{\left[-{6},{2}\right]}.$

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Wednesday, August 31, 2016

$\displaystyle{4}{x}+{{y}}^{{2}}={12},{x}={y}$ Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $\displaystyle{x}$ or $\displaystyle{y}$ . Draw a typical...

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How does author Elie Wiesel use symbolism to contribute to the meaning of Night?

Wednesday, August 31, 2016

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to or . Draw a typical...

No comments:

Post a Comment

How does author Elie Wiesel use symbolism to contribute to the meaning of Night?

$\displaystyle{4}{x}+{{y}}^{{2}}={12},{x}={y}$ Sketch the region enclosed by the given curves. Decide whether to integrate with respect to $\displaystyle{x}$ or $\displaystyle{y}$ . Draw a typical...