Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 82872 by abdomathmax last updated on 25/Feb/20

1)find ∫∫_W  ((xdx)/(a^2  +x^2  +y^2 )) with  W_a →x^2  +y^2  ≤a^2  and x>0     (a>0)  2)calculate ∫∫_W_1     ((xdx)/(x^2 +y^2  +1))

$$\left.\mathrm{1}\right){find}\:\int\int_{{W}} \:\frac{{xdx}}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\:{with} \\ $$ $${W}_{{a}} \rightarrow{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant{a}^{\mathrm{2}} \:{and}\:{x}>\mathrm{0}\:\:\:\:\:\left({a}>\mathrm{0}\right) \\ $$ $$\left.\mathrm{2}\right){calculate}\:\int\int_{{W}_{\mathrm{1}} } \:\:\:\frac{{xdx}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:+\mathrm{1}} \\ $$

Commented bymathmax by abdo last updated on 25/Feb/20

1)  we use the diffeomorphism  (r,θ)→(x,y)/x=rcosθ  and y=rsinθ  wehave  x^2  +y^2 ≤a^2  ⇒0≤r≤a   x>0 ⇒θ∈]−(π/2),(π/2)[  ⇒∫∫_W_a   ((xdxdy)/(a^2  +r^2 )) =∫_0 ^a  ∫_(−(π/2)) ^(π/2)   ((rcosθ rdr dθ)/(a^2  +r^2 ))   ∫_0 ^a  (r^2 /(r^2  +a^2 )) dr ×∫_(−(π/2)) ^(π/2)  cosθ dθ  =2 ∫_0 ^a  (r^2 /(r^2  +a^2 ))dr  =2 ∫_0 ^a  ((r^2  +a^2 −a^2 )/(r^2  +a^2 ))dr =2a −2a^2  ∫_0 ^a  (dr/(r^2  +a^2 ))  (ch. r=ax)  =2a−2a^2  ∫_0 ^1  ((adx)/(a^2 (1+x^2 ))) =2a−2a [arctanx]_0 ^1   =2a−2a×(π/4) =(2−(π/2))a   2) ∫∫_W_1      ((xdx)/(x^2  +y^2  +1)) =_(a=1)    2−(π/2)

$$\left.\mathrm{1}\right)\:\:{we}\:{use}\:{the}\:{diffeomorphism}\:\:\left({r},\theta\right)\rightarrow\left({x},{y}\right)/{x}={rcos}\theta \\ $$ $$\left.{and}\:{y}={rsin}\theta\:\:{wehave}\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} \:\Rightarrow\mathrm{0}\leqslant{r}\leqslant{a}\:\:\:{x}>\mathrm{0}\:\Rightarrow\theta\in\right]−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\left[\right. \\ $$ $$\Rightarrow\int\int_{{W}_{{a}} } \:\frac{{xdxdy}}{{a}^{\mathrm{2}} \:+{r}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{{a}} \:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{rcos}\theta\:{rdr}\:{d}\theta}{{a}^{\mathrm{2}} \:+{r}^{\mathrm{2}} } \\ $$ $$\:\int_{\mathrm{0}} ^{{a}} \:\frac{{r}^{\mathrm{2}} }{{r}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:{dr}\:×\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta\:{d}\theta\:\:=\mathrm{2}\:\int_{\mathrm{0}} ^{{a}} \:\frac{{r}^{\mathrm{2}} }{{r}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dr} \\ $$ $$=\mathrm{2}\:\int_{\mathrm{0}} ^{{a}} \:\frac{{r}^{\mathrm{2}} \:+{a}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{r}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dr}\:=\mathrm{2}{a}\:−\mathrm{2}{a}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{{a}} \:\frac{{dr}}{{r}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:\:\left({ch}.\:{r}={ax}\right) \\ $$ $$=\mathrm{2}{a}−\mathrm{2}{a}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{adx}}{{a}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\mathrm{2}{a}−\mathrm{2}{a}\:\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$ $$=\mathrm{2}{a}−\mathrm{2}{a}×\frac{\pi}{\mathrm{4}}\:=\left(\mathrm{2}−\frac{\pi}{\mathrm{2}}\right){a}\: \\ $$ $$\left.\mathrm{2}\right)\:\int\int_{{W}_{\mathrm{1}} } \:\:\:\:\frac{{xdx}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+\mathrm{1}}\:=_{{a}=\mathrm{1}} \:\:\:\mathrm{2}−\frac{\pi}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com