Surface Integral From Div, Grad, Curl and all that

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The discussion revolves around evaluating a surface integral using the divergence theorem, specifically for the vector field F(x,y,z) = j*y + k. The user is struggling to match their calculated result with the book's answer, which is π/2. They attempted to convert the integral into polar coordinates but ended up with a different value. Other participants pointed out that the user may have overlooked contributions from the bottom surface of the integral and clarified that the correct answer should actually be 3π/2. The conversation highlights common pitfalls in surface integral calculations and emphasizes the importance of considering all surfaces involved.
SoCoToAh
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First, this is my first time actually posting anything so hi PF!Second, I have been working out of Div, Grad, Curl and all that. This problem has me stumped for some reason. My answer never comes out to be the same as the books. If you could help me figure out where I am going wrong I would greatly appreciate it.

Homework Statement


Evaluate the surface integral using the following equation.

Homework Equations


\int\int\vec{F}(x,y,z)\hat{n}=\int\int-F_{x}∂S/∂x-F_{y}∂S/∂y+F_{z}dS

Where,
S(x,y,f(x,y))=z=1-x^{2}-y^{2}

above the xy plane

and

\vec{F}(x,y,z)=\hat{j}y+\hat{k}

The Attempt at a Solution


After setting up the integral I attempt to switch to polar:

\int\int-F_{x}∂S/∂x-F_{y}∂S/∂y+F_{z}dS=\int\int2y^{2}+1 dxdy

=\int\int2((rSin[θ])^{2}+1)rdrdθ
from r=0 to r=1 and θ=0 to θ=2\pi
=\intSin^{2}[θ]/2dθ+\pi

\intSin^{2}[θ]dθ=\int(1-Cos[2θ])/2dθ
=θ/2-Sin[2θ]/4

\intSin^{2}[θ]/2dθ+\pi=\pi/4+\pi

The book says the answer is \pi/2. Looking at what I have I I could say 1=r^{2}
Then \int\int2((rSin[θ])^{2}+1)rdrdθ=\int\int2((rSin[θ])^{2}+r^{2})rdrdθ and thus equal to \pi/2.

Where am I going wrong? Thanks in advance.
 
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SoCoToAh said:
First, this is my first time actually posting anything so hi PF!Second, I have been working out of Div, Grad, Curl and all that. This problem has me stumped for some reason. My answer never comes out to be the same as the books. If you could help me figure out where I am going wrong I would greatly appreciate it.

Homework Statement


Evaluate the surface integral using the following equation.

Homework Equations


\int\int\vec{F}(x,y,z)\hat{n}=\int\int-F_{x}∂S/∂x-F_{y}∂S/∂y+F_{z}dS

Where,
S(x,y,f(x,y))=z=1-x^{2}-y^{2}

above the xy plane

and

\vec{F}(x,y,z)=\hat{j}y+\hat{k}

The Attempt at a Solution


After setting up the integral I attempt to switch to polar:

\int\int-F_{x}∂S/∂x-F_{y}∂S/∂y+F_{z}dS=\int\int2y^{2}+1 dxdy

=\int\int2((rSin[θ])^{2}+1)rdrdθ
from r=0 to r=1 and θ=0 to θ=2\pi
=\intSin^{2}[θ]/2dθ+\pi

\intSin^{2}[θ]dθ=\int(1-Cos[2θ])/2dθ
=θ/2-Sin[2θ]/4

\intSin^{2}[θ]/2dθ+\pi=\pi/4+\pi
You should get ##3\pi/2##. The first term works out to ##\pi/2##, not ##\pi/4##.

The book says the answer is \pi/2. Looking at what I have I I could say 1=r^{2}
Then \int\int2((rSin[θ])^{2}+1)rdrdθ=\int\int2((rSin[θ])^{2}+r^{2})rdrdθ and thus equal to \pi/2.

Where am I going wrong? Thanks in advance.
To get an answer of ##\pi/2##, you need to include the contribution from the bottom surface. What exactly did the problem ask you to calculate?
 
Actually, after looking, it appears I may have looked at the wrong answer. 3(pi)/2 is the answer. Sorry for my carelessness.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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