Vector Calculus Questions

Question Number 100074 by kungmikami last updated on 24/Jun/20

Answered by smridha last updated on 25/Jun/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\int_{\mathrm{0}} ^{\sqrt{\boldsymbol{{y}}}} \left(\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}+\boldsymbol{{xy}}^{\mathrm{2}} \right)\boldsymbol{{dx}}\right]\boldsymbol{{dy}} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{dy}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}+\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\sqrt{{y}}} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}}^{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{y}}^{\mathrm{3}} \right]\boldsymbol{{dy}} \\$$$$=\left[\frac{\mathrm{1}}{\mathrm{3}}.\frac{\boldsymbol{{y}}^{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{8}}\boldsymbol{{y}}^{\mathrm{4}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$=\frac{\mathrm{2}}{\mathrm{21}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{37}}{\mathrm{168}}\left(\boldsymbol{{unit}}\right)^{\mathrm{2}} \\$$