Question : Find curl of the vector F=(x^2-y^2)i-(2xy+y)j Solution : Given vector \[\vec{F}=\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\] Then curl of F is \[ curl\left(\vec{F}\right)=\nabla\times\vec{F}=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)\times\left(\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\right)\] \[=\left(\hat{i}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial x}-\left(\hat{i}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial x}+\left(\hat{j}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial y}-\left(\hat{j}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial y}\] Since \[\hat{i}\times\hat{i}=\hat{j}\times\hat{j}=0\ ,\ \hat{i}\times\hat{j}=\hat{k},\ \hat{j}\times\hat{i}=-\hat{k}\] \[curl\left(\vec{F}\right)=0-\left(\hat{i}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial x}+\left(\hat{j}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial y}-0 curl\left(\vec{F}\right)=0-\hat{k}\left(2y\right)-\hat{k}\left(-2y\right)-0\] \[curl\left(\vec{F}\right)=-2y\hat{k}-2y\hat{k}=\vec{0}\] is the required…
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