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Question Number 98520 by bobhans last updated on 14/Jun/20

$$\mathrm{Integrate}\mathrm{the}\mathrm{function}\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{xy}({\mathrm{x}}^2+{\mathrm{y}}^2)$$$$\mathrm{over}\mathrm{the}\mathrm{domain}\mathrm{R}:\{-3⩽{\mathrm{x}}^2-{\mathrm{y}}^2⩽3,1⩽\mathrm{xy}⩽4\}$$

Answered by john santu last updated on 14/Jun/20

$$\mathrm{I}=\int \underset{\mathrm{R}}{\int }xy(x^2+y^2)dxdy$$$$put{x}^2-y^2=u;xy=v$$$$limits:-3⩽u⩽3;1⩽v⩽4\mathrm{which}$$$$\mathrm{bounds}\mathrm{region}\mathrm{R}$$$$\mathrm{J}=\frac{\mathrm{J}(\mathrm{u},\mathrm{v})}{\mathrm{J}(\mathrm{x},\mathrm{y})}=\left|\begin{array}{c}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}\\ \frac{\partial \mathrm{v}}{\partial \mathrm{x}}\frac{\partial \mathrm{v}}{\partial \mathrm{y}}\end{array}\right|=\left|\begin{array}{c}2\mathrm{x}-2\mathrm{y}\\ \mathrm{y}\mathrm{x}\end{array}\right|$$$$={2\mathrm{x}}^2+{2\mathrm{y}}^2$$$$\mathrm{J}(\mathrm{x},\mathrm{y})=\frac{\mathrm{J}(\mathrm{u},\mathrm{v})}{2({\mathrm{x}}^2+{\mathrm{y}}^2)}$$$$\mathrm{I}=\frac{1}{2}\underset{\mathrm{u}=-3}{\stackrel{\mathrm{u}=3}{\int }}\underset{\mathrm{v}=1}{\stackrel{\mathrm{v}=4}{\int }}\mathrm{v}\mathrm{dvdu}$$$${=\frac{1}{2}[3-(-3))\left(\frac{{\mathrm{v}}^2}{2}\right)]}_1^4$$$$=3\left(\frac{15}{2}\right)=\frac{45}{2}◼$$

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