Question Number 129816 by BHOOPENDRA last updated on 19/Jan/21
Answered by mathmax by abdo last updated on 19/Jan/21
![∫∫∫_C_f z e^(x^2 +y^2 ) dxdydz =_ { ((x=rcosθ)),((y=rsinθ)) :} ∫_2 ^3 ( ∫_0 ^2 e^r^2 rdr∫_0 ^(2π) dθ)zdz =2π[(1/2)e^r^2 ]_0 ^2 ×[(z^2 /2)]_2 ^3 =π(e^4 −1)((9/2)−2) =π(e^4 −1)×(5/2) =((5π)/2)(e^4 −1)](Q129820.png)
$$\int\int\int_{\mathrm{C}_{\mathrm{f}} } \mathrm{z}\:\mathrm{e}^{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \mathrm{dxdydz}\:=_{\begin{cases}{\mathrm{x}=\mathrm{rcos}\theta}\\{\mathrm{y}=\mathrm{rsin}\theta}\end{cases}} \:\:\:\int_{\mathrm{2}} ^{\mathrm{3}} \left(\:\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{e}^{\mathrm{r}^{\mathrm{2}} } \mathrm{rdr}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{d}\theta\right)\mathrm{zdz} \\ $$$$=\mathrm{2}\pi\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{\mathrm{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{2}} ×\left[\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{2}} ^{\mathrm{3}} \:=\pi\left(\mathrm{e}^{\mathrm{4}} −\mathrm{1}\right)\left(\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{2}\right)\:=\pi\left(\mathrm{e}^{\mathrm{4}} −\mathrm{1}\right)×\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{5}\pi}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{4}} −\mathrm{1}\right) \\ $$
Commented by BHOOPENDRA last updated on 19/Jan/21
$${tq}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 19/Jan/21
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}. \\ $$