Notes: `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...

Tuesday, 16 June 2015

$\displaystyle{4}{x}+{{y}}^{{2}}={12},{x}={y}$ 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...

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 $\displaystyle{x}\in{\left[-{6},{2}\right]}.$

Notes

Tuesday, 16 June 2015

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

No comments:

Post a Comment

Is there any personification in "The Tell-Tale Heart"?

Tuesday, 16 June 2015

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

No comments:

Post a Comment

Is there any personification in &quot;The Tell-Tale Heart&quot;?

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

Is there any personification in "The Tell-Tale Heart"?