Notes: `y = sin(x), y = cos(x), 0

Tuesday, 24 October 2017

$\displaystyle{y}={\sin{{\left({x}\right)}}},{y}={\cos{{\left({x}\right)}}},{0}$

We use Washer method to evaluate the volume of the solid.

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

Here

$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}},{g{{\left({x}\right)}}}={\cos{{x}}}$

$\displaystyle{x}={0}{\quad\text{and}\quad}\frac{\pi}{{4}}$

about $\displaystyle{y}=-{1}$

$V = pi int_0^(pi/4) {[(sinx -(-1)]^2 - [cosx-(-1)]^2}dx$

$V = pi int_0^(pi/4) {[(sinx + 1]^2 - [cosx + 1]^2}dx$

$\displaystyle{V}=\pi{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{{\sin}}^{{2}}{x}+{2}{\sin{{x}}}\ldots\right.}$

We use Washer method to evaluate the volume of the solid.

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

Here

$\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}},{g{{\left({x}\right)}}}={\cos{{x}}}$

$\displaystyle{x}={0}{\quad\text{and}\quad}\frac{\pi}{{4}}$

about $\displaystyle{y}=-{1}$

$V = pi int_0^(pi/4) {[(sinx -(-1)]^2 - [cosx-(-1)]^2}dx$

$V = pi int_0^(pi/4) {[(sinx + 1]^2 - [cosx + 1]^2}dx$

$\displaystyle{V}=\pi{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{{\sin}}^{{2}}{x}+{2}{\sin{{x}}}+{1}-{{\cos{{x}}}}^{{2}}-{2}{\cos{{x}}}-{1}\right]}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{{\sin}}^{{2}}{x}+{2}{\sin{{x}}}-{{\cos}}^{{2}}{x}-{2}{\cos{{x}}}\right]}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{2}{\sin{{x}}}-{2}{\cos{{x}}}-{\left({{\cos}}^{{2}}{x}+{{\sin}}^{{2}}{x}\right)}\right]}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\left[{2}{\sin{{x}}}-{2}{\cos{{x}}}-{\cos{{2}}}{x}\right]}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi{\left[{\int_{{0}}^{{\frac{\pi}{{4}}}}}{2}{\sin{{x}}}{\left.{d}{x}\right.}-{\int_{{0}}^{{\frac{\pi}{{4}}}}}{2}{\cos{{x}}}{\left.{d}{x}\right.}-{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\cos{{2}}}{x}{\left.{d}{x}\right.}\right]}$

$\displaystyle{V}=\pi{\left[-{2}{\cos{{x}}}-{2}{\sin{{x}}}-\frac{{1}}{{2}}{\sin{{2}}}{x}\right]}{{\mid}_{{0}}^{{\frac{\pi}{{4}}}}}$

$\displaystyle{V}=\pi{\left[-{2}{\cos{{\left(\frac{\pi}{{4}}\right)}}}-{2}{\sin{{\left(\frac{\pi}{{4}}\right)}}}-\frac{{1}}{{2}}{\sin{{2}}}\cdot{\left(\frac{\pi}{{4}}\right)}-{\left(-{2}{\cos{{0}}}-{2}{\sin{{0}}}-\frac{{1}}{{2}}{\sin{{2}}}\cdot{0}\right)}\right]}$