Question

Evaluate the iterated integral: a)int_0^3 int_0^(pi/3) int_0^4 r xx cos theta drd theta dz b)int_0^(pi/2) int_0^3 int_0^(4 - z) zdrdzd theta

Integrals
ANSWERED
asked 2021-09-02

Evaluate the iterated integral:
a) \(\displaystyle{\int_{{0}}^{{3}}}{\int_{{0}}^{{\frac{\pi}{{3}}}}}{\int_{{0}}^{{4}}}{r}\times \cos{\theta}{d}{r}{d}\theta{\left.{d}{z}\right.}\)

b) \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{3}}}{\int_{{0}}^{{{4}-{z}}}}{z}{d}{r}{\left.{d}{z}\right.}{d}\theta\)

c) \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{2}}}\rho^{2}{d}\rho{d}\theta{d}\phi\)

d) \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\int_{{0}}^{{ \cos{{\left(\phi\right)}}}}} \cos{\theta}{d}\rho{d}\phi{d}\theta\)

Expert Answers (1)

2021-09-03

a) \(\displaystyle{\int_{{0}}^{{3}}}{\int_{{0}}^{{\frac{\pi}{{3}}}}}{\int_{{0}}^{{4}}}{r}\times \cos{\theta}{d}{r}{d}\theta{\left.{d}{z}\right.}=\)

\(\displaystyle={\int_{{0}}^{{3}}}{\int_{{0}}^{{\frac{\pi}{{3}}}}} \cos{\theta}{{\left[\gamma^{{\frac{2}{{2}}}}\right]}_{{0}}^{{4}}}{d}\theta{\left.{d}{z}\right.}=\)

\(\displaystyle={\int_{{0}}^{{3}}}{\int_{{0}}^{{\frac{\pi}{{3}}}}}{8} \cos{\theta}\times{d}\theta\times{\left.{d}{z}\right.}=\)

\(\displaystyle={\int_{{0}}^{{3}}}{8}\times{{\left[ \sin{\theta}\right]}_{{0}}^{{\frac{\pi}{{3}}}}}{\left.{d}{z}\right.}=\)

\(\displaystyle={\int_{{0}}^{{3}}}{8}\times\frac{\sqrt{{{3}}}}{{2}}{\left.{d}{z}\right.}=\)

\(\displaystyle={4}\sqrt{{{3}}}{\int_{{0}}^{{3}}}{\left.{d}{z}\right.}=\)

\(\displaystyle={4}\sqrt{{{3}}}{{\left[{z}\right]}_{{0}}^{{3}}}=\)

\(\displaystyle={12}\sqrt{{{3}}}\)

b) \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{3}}}{\int_{{0}}^{{{4}-{z}}}}{z}{d}{r}{\left.{d}{z}\right.}{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{3}}}{z}\times{{\left[\gamma\right]}_{{0}}^{{{4}-{z}}}}{\left.{d}{z}\right.}\times{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{3}}}{z}\times{\left[{\left({4}–{2}\right)}–{0}\right]}{\left.{d}{z}\right.}{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{3}}}{4}{z}–{z}^{2}\times{\left.{d}{z}\right.}{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left[{\left({4}\frac{{z}^{2}}{{2}}\right)}–\frac{{z}^{3}}{{3}}\right]}_{{0}}^{{3}}}{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({18}-{9}\right)}\times{d}\theta=\)

\(\displaystyle={9}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{d}\theta=\)

\(\displaystyle={9}{{\left[\theta\right]}_{{0}}^{{\frac{\pi}{{2}}}}}=\)

\(\displaystyle=\frac{{{9}\pi}}{{2}}\)

c) \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{2}}}\rho^{2}{d}\rho{d}\theta{d}\phi=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left[\rho^{{\frac{3}{{3}}}}\right]}_{{0}}^{{2}}}{d}\theta{d}\phi=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{8}{{3}}{d}\theta{d}\phi=\)

\(\displaystyle=\frac{8}{{3}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left[\theta\right]}_{{0}}^{{\frac{\pi}{{2}}}}}{d}\phi=\)

\(\displaystyle=\frac{8}{{3}}\times\frac{\pi}{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{d}\phi=\)

\(\displaystyle=\frac{8}{{3}}\times\frac{\pi}{{2}}{{\left[\phi\right]}_{{0}}^{{\frac{\pi}{{2}}}}}=\)

\(\displaystyle=\frac{8}{{3}}\times\frac{\pi}{{2}}\times\frac{\pi}{{2}}=\)

\(\displaystyle=\frac{{{2}\pi^{2}}}{{3}}\)

d) \(\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\int_{{0}}^{{\frac{\pi}{{4}}}}}{\int_{{0}}^{{ \cos{{\left(\phi\right)}}}}} \cos{\theta}{d}\rho{d}\phi{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{\int_{{0}}^{{\frac{\pi}{{4}}}}} \cos{\theta}{{\left[\rho\right]}_{{0}}^{ \cos{\phi}}}{d}\phi{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}}{\int_{{0}}^{{\frac{\pi}{{4}}}}} \cos{\theta} \cos{\phi}\times{d}\phi{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{4}}}}} \cos{\theta}{{\left[ \sin{\phi}\right]}_{{0}}^{{\frac{\pi}{{4}}}}}{d}\theta=\)

\(\displaystyle={\int_{{0}}^{{\frac{\pi}{{2}}}}}\frac{1}{\sqrt{{2}}} \cos{\theta}\times{d}\theta=\)

\(\displaystyle=\frac{1}{\sqrt{{2}}}{{\left[ \sin{\theta}\right]}_{{0}}^{{\frac{\pi}{{4}}}}}=\)

\(\displaystyle=\frac{1}{\sqrt{{2}}}\times\frac{1}{\sqrt{{2}}}=\)


\(=\displaystyle\frac{1}{{2}}\)

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