SLD
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Integrate the function x^2+3y over the circle of radius 1 centered at y=1, x=0.
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barbos
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SLD
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How?
lpetrich
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I will do it.
First, change to polar coordinates centered on the disk: x = r*cos(a), y = r*sin(a) + 1
Radius: r, angle: a
Substitute into the integrand: x^2+3*y = r^2*cos(a)^2 + 3*(r*sin(a) + 1)
The area integration measure is r*d(r)*d(a) over r = 0 to 1 and a = 0 to 2*pi
First integrate over the radius: (1/4)*cos(a)^2 + sin(a) + (3/2)
Now integrate over the angle: (1/4)*pi + 3/2*(2*pi) = (13/4)*pi
I checked my work with Mathematica.
First, change to polar coordinates centered on the disk: x = r*cos(a), y = r*sin(a) + 1
Radius: r, angle: a
Substitute into the integrand: x^2+3*y = r^2*cos(a)^2 + 3*(r*sin(a) + 1)
The area integration measure is r*d(r)*d(a) over r = 0 to 1 and a = 0 to 2*pi
First integrate over the radius: (1/4)*cos(a)^2 + sin(a) + (3/2)
Now integrate over the angle: (1/4)*pi + 3/2*(2*pi) = (13/4)*pi
I checked my work with Mathematica.
steve_bank
Diabetic retinopathy and poor eyesight. Typos ...
Calculus 101
You have to find the limits of integration from the intercepts. Find the points where f1() and f2() coincide. Solve for f1() = f2();
You can do a pencil and paper sketch to get a s handle on it.
One method is to subtract one area from the other. A little easier to see is what is the area of y = e^over the line y = 5 for x = 0 to 100?
Sketch a graph of the two on each other. The area of the line is y*x. The integral of e^x is e^hilimit - e^lolimit.
Sketch the functions on graph paper nd count the squares. That is what it reduces to.
You have to find the limits of integration from the intercepts. Find the points where f1() and f2() coincide. Solve for f1() = f2();
You can do a pencil and paper sketch to get a s handle on it.
One method is to subtract one area from the other. A little easier to see is what is the area of y = e^over the line y = 5 for x = 0 to 100?
Sketch a graph of the two on each other. The area of the line is y*x. The integral of e^x is e^hilimit - e^lolimit.
Sketch the functions on graph paper nd count the squares. That is what it reduces to.
Bomb#20
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It's also helpful to sanity-check this sort of problem by bounding it. Integration is linear, so it's the same problem as these two:
Integrate the function 3y over the circle of radius 1 centered at y=1, x=0.
Integrate the function x^2 over the circle of radius 1 centered at y=1, x=0.
The first problem is simply the volume of a cylinder with a diagonal upper surface, z=0 at (0,0) and z=6 at (0,2). The slant of the upper surface has no effect on the volume since the extra volume at y>1 exactly cancels the missing volume at y < 1, so this is simply the volume of a cylinder with radius 1 and height 3, that is, 3 pi.
The second problem is the volume of a cylinder with a parabolic slice taken out, with maximum height 1 at (-1,1) and (1,1). So its volume has to be between 0 and pi. So the answer to the original problem must be somewhere between 3 pi and 4 pi.
barbos
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Wait, it's 2D integral?
over a circle means a line.
Regardless, since I integrated over a circle (1D) and it is Pi*(6*r + r^3). all I need to do is to integrate over radius [0,1]
which is Pi*(3+1./4)
You should have said "inside a circle" instead of "over a circle", or "over a disk"
over a circle means a line.
Regardless, since I integrated over a circle (1D) and it is Pi*(6*r + r^3). all I need to do is to integrate over radius [0,1]
which is Pi*(3+1./4)
You should have said "inside a circle" instead of "over a circle", or "over a disk"
SLD
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Yes. Disk. My bad.