Notes: `y = x^2, x = y^2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch...

Saturday, 14 January 2017

$\displaystyle{y}={{x}}^{{2}},{x}={{y}}^{{2}}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch...

The volume of the solid obtained by rotating the region bounded by the curves $\displaystyle{{y}}^{{2}}={x}$ and $\displaystyle{y}={{x}}^{{2}}$ about y = 1, can be evaluated using the washer method, such that:

$\displaystyle{V}={\int_{{a}}^{{b}}}\pi\cdot{\left({{f}}^{{2}}{\left({x}\right)}-{{g}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

You need to find the endpoint of interval, hence, you need to solve for x the following equation, such that:

$\displaystyle\sqrt{{x}}={{x}}^{{2}}\Rightarrow{x}-{{x}}^{{4}}={0}\Rightarrow{x}{\left({1}-{{x}}^{{3}}\right)}={0}\Rightarrow{x}=\ldots$

$\displaystyle{V}={\int_{{a}}^{{b}}}\pi\cdot{\left({{f}}^{{2}}{\left({x}\right)}-{{g}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}$

You need to find the endpoint of interval, hence, you need to solve for x the following equation, such that:

$\displaystyle\sqrt{{x}}={{x}}^{{2}}\Rightarrow{x}-{{x}}^{{4}}={0}\Rightarrow{x}{\left({1}-{{x}}^{{3}}\right)}={0}\Rightarrow{x}={0}$ and $\displaystyle{x}={1}$

You need to notice that $\displaystyle{{x}}^{{2}}<\sqrt{{x}}$ on [0,1], such that:

$\displaystyle{V}={\int_{{0}}^{{1}}}\pi\cdot{\left({{\left({\left(\sqrt{{x}}\right)}-{1}\right)}}^{{2}}-{{\left({{x}}^{{2}}-{1}\right)}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi\cdot{\int_{{0}}^{{1}}}{\left({x}-{2}\sqrt{{x}}+{1}\right)}{\left.{d}{x}\right.}-\pi\cdot{\int_{{0}}^{{1}}}{\left({{x}}^{{4}}-{2}{{x}}^{{2}}+{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={\left(\pi\cdot\frac{{{x}}^{{2}}}{{2}}-{\left(\frac{{4}}{{3}}\right)}\pi\cdot{{x}}^{{\frac{{3}}{{2}}}}+\pi\cdot{x}-\pi\cdot\frac{{{x}}^{{5}}}{{5}}+{2}\pi\cdot\frac{{{x}}^{{3}}}{{3}}-\pi\cdot{x}\right)}{{\mid}_{{0}}^{{1}}}$

$\displaystyle{V}=\pi\cdot\frac{{{1}}^{{2}}}{{2}}-{\left(\frac{{4}}{{3}}\right)}\pi\cdot{{1}}^{{\frac{{3}}{{2}}}}-\pi\cdot\frac{{{1}}^{{5}}}{{5}}+{2}\pi\cdot\frac{{{1}}^{{3}}}{{3}}-{0}$

$\displaystyle{V}=\frac{\pi}{{2}}+\frac{{{4}\pi}}{{3}}-\frac{\pi}{{5}}-\frac{{{2}\pi}}{{3}}$

$\displaystyle{V}=\frac{\pi}{{2}}+\frac{{{2}\pi}}{{3}}-\frac{\pi}{{5}}$

$\displaystyle{V}=\frac{{{15}\pi+{20}\pi-{6}\pi}}{{30}}$

$\displaystyle{V}=\frac{{{29}\pi}}{{30}}$

Hence, evaluating the volume of the solid obtained by rotating the region bounded by the curves $\displaystyle{{y}}^{{2}}={x},{y}={{x}}^{{2}}$ about y = 1 , using the washer method, yields $\displaystyle{V}=\frac{{{29}\pi}}{{30}}.$