# the lower limit is 0 the upper limit is x on the first and thirdintegral. the second integral lower limit is 0 andupper limit is 2x. int sin theta dtheta int d psi int r^2 dr=?

the lower limit is 0 the upper limit is x on the first and thirdintegral. the second integral lower limit is 0 andupper limit is 2x.
$\int \mathrm{sin}\theta d\theta \int d\psi \int {r}^{2}﻿dr$
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Jazlene Dickson
${\int }_{0}^{x}{\int }_{0}^{2x}{\int }_{0}^{x}\mathrm{sin}\theta {r}^{2}drd\psi d\theta$
$={\int }_{0}^{x}{\int }_{0}^{2x}\frac{1}{3}{x}^{3}\mathrm{sin}\theta d\psi d\theta ={\int }_{0}^{x}\left(\frac{2}{3}{x}^{4}\right)\mathrm{sin}\theta d\theta$
$=\frac{2}{3}{x}^{4}\left[-\mathrm{cos}\theta \right]{|}_{0}^{x}=\frac{2{x}^{4}}{3}\left[-\mathrm{cos}x+\mathrm{cos}\left(0\right)\right]$
$=\frac{2{x}^{4}}{3}\left[1-\mathrm{cos}\left(x\right)\right]$
###### Not exactly what you’re looking for?
vangstosiis
$\int \mathrm{sin}\theta d\theta \int d\psi \int {r}^{2}dr={\int }_{0}^{2}{\int }_{0}^{2x}\int {r}^{2}drd\psi d\theta =\frac{2{r}^{3}{x}^{2}}{3}$