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If f(x)=1, then the derivative of f(x) with respect to x (denoted as f′(x) or dxdf) is zero because the function is constant.
f′(x)=dxd(1)=0
The integral of f(x)=1 with respect to x (denoted as ∫f(x)dx) is simply x plus a constant of integration (often denoted as C).
∫1dx=x+C
So, in summary:
f′(x)=0
∫1dx=x+C
where C is the constant of integration.
When changing variables from Cartesian coordinates (x, y) to polar coordinates (r, θ) in a double integral, the limits of integration need to be adjusted accordingly. If the original limits for x and y extend from negative infinity to positive infinity, it typically implies that you are integrating over the entire plane. In polar coordinates, the corresponding region would be described using the radial coordinate (r) and the angular coordinate (θ).
Let's go through the process step by step:
Original double integral:
∫−∞∞∫−∞∞f(x,y)dxdy
Now, let's express the Cartesian coordinates (x, y) in terms of polar coordinates (r, θ):
x=rcos(θ) y=rsin(θ)
Next, we need to express the differential area element dxdy in terms of polar coordinates. The area element in Cartesian coordinates is dxdy, and in polar coordinates, it becomes rdrdθ. Therefore:
dxdy=rdrdθ
Now, substitute these expressions into the original integral:
∫−∞∞∫−∞∞f(x,y)dxdy=∫02π∫0∞f(rcos(θ),rsin(θ))rdrdθ
In polar coordinates, the limits for r typically start from 0 to ∞, and the limits for θ range from 0 to 2π to cover the entire plane.
So, when changing variables from x, y to r, θ in a double integral where the original limits are from −∞ to +∞, the new limits become:
∫−∞∞∫−∞∞f(x,y)dxdy
becomes
∫02π∫0∞f(rcos(θ),rsin(θ))rdrdθ
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