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Question Number 170545 by kndramaths last updated on 26/May/22
    please help me to find this.     a= ∫∫_D ((ydxdy)/(a^2 +x^2 )) D:{x≥0.y≥0.x^2 +y^2 ≤a^2 }     b=∫∫∫_v (x−y+z)^2 dxdydz    v:{x=0.y=0.z=0 x+z=1.y+z=1}      c=∫∫∫_V xydxdydz          V:{0≤z≤1. x^2 +y^2 ≤z^2 }
$$\:\:\:\:{please}\:{help}\:{me}\:{to}\:{find}\:{this}. \\ $$$$\:\:\:{a}=\:\int\int_{{D}} \frac{{ydxdy}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{D}:\left\{{x}\geqslant\mathrm{0}.{y}\geqslant\mathrm{0}.{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} \right\} \\ $$$$\:\:\:{b}=\int\int\int_{{v}} \left({x}−{y}+{z}\right)^{\mathrm{2}} {dxdydz} \\ $$$$\:\:{v}:\left\{{x}=\mathrm{0}.{y}=\mathrm{0}.{z}=\mathrm{0}\:{x}+{z}=\mathrm{1}.{y}+{z}=\mathrm{1}\right\} \\ $$$$\:\:\:\:{c}=\int\int\int_{{V}} {xydxdydz} \\ $$$$\:\:\:\:\:\:\:\:{V}:\left\{\mathrm{0}\leqslant{z}\leqslant\mathrm{1}.\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{z}^{\mathrm{2}} \right\} \\ $$
Commented by kndramaths last updated on 27/May/22
    please help me to find this.     a= ∫∫_D ((ydxdy)/(a^2 +x^2 )) D:{x≥0.y≥0.x^2 +y^2 ≤a^2 }     b=∫∫∫_v (x−y+z)^2 dxdydz    v:{x=0.y=0.z=0 x+z=1.y+z=1}      c=∫∫∫_V xydxdydz          V:{0≤z≤1. x^2 +y^2 ≤z^2 }
$$\:\:\:\:{please}\:{help}\:{me}\:{to}\:{find}\:{this}. \\ $$$$\:\:\:{a}=\:\int\int_{{D}} \frac{{ydxdy}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{D}:\left\{{x}\geqslant\mathrm{0}.{y}\geqslant\mathrm{0}.{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} \right\} \\ $$$$\:\:\:{b}=\int\int\int_{{v}} \left({x}−{y}+{z}\right)^{\mathrm{2}} {dxdydz} \\ $$$$\:\:{v}:\left\{{x}=\mathrm{0}.{y}=\mathrm{0}.{z}=\mathrm{0}\:{x}+{z}=\mathrm{1}.{y}+{z}=\mathrm{1}\right\} \\ $$$$\:\:\:\:{c}=\int\int\int_{{V}} {xydxdydz} \\ $$$$\:\:\:\:\:\:\:\:{V}:\left\{\mathrm{0}\leqslant{z}\leqslant\mathrm{1}.\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{z}^{\mathrm{2}} \right\} \\ $$

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