Sorry if my title was vague but i was not entirely sure what its called.
Anyways i was solving some work and energy problems and encountered this integration:
$$\int_{2,1,4}^{2,-3,3} 2x\sin^2y \,dx+ (z^2+x^2\sin2y)\,dy+ 2zy\,dz$$
Could someone explain how to solve similar integrals? Is there something to do with partial derivatives?
The book gave the answer saying it was of form
$$\int_{2,1,4}^{2,-3,3} d(x^2\sin^2y + z^2y)$$
I cant quite understand how that integral came.
In some cases we can split up the integral into $dx$ and $dy$ and integrate separately. Is it possible to do here?
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$\begingroup$You should have stated that the field that does work on the object is conservative, and if it is then differential amounts of work done can be written as $dF$ where it is presented to you in an expanded form like on the right hand side below.
$$dF(x,y,z)=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz$$
We need to match the terms above with that given in the problem and find F.
Then equating terms
$$\frac{\partial F}{\partial z}=2zy$$
Integrating
$$F(x,y,z)=yz^2+h(x,y)$$
$$\frac{\partial F}{\partial y}=z^2+\frac{\partial h}{\partial y}=z^2+x^2sin(2y)$$
Integrating we find
$$h(x,y)=-\frac{1}{2}x^2cos(2y)+g(x)$$
$$F(x,y,z)=yz^2-\frac{1}{2}x^2cos(2y)+g(x)$$
$$\frac{\partial F}{\partial x}=-xcos(2y)+g'(x)=2xsin^2(y)$$
$$g'(x)=x(2sin^2(y)+cos(2y))$$
Using a double angle formula
$$g'(x)=x(2sin^2(y)+cos^2(y)-sin^2(y))=x(sin^2(y)+cos^2(y))=x$$
$$g'(x)=x$$
$$g(x)=x^2/2$$
$$F(x,y,z)=yz^2-\frac{1}{2}x^2cos(2y)+x^2/2$$
Recognizing $1-Cos(2y)=1-Cos^2(y)+Sin^2(y)=2Sin^2(y)$ and substituting you get
$$F(x,y,z)=yz^2+x^2sin^2(y)$$
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