Pertemuan
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April 22, 2023 Kalkulus II 1
PertemPertemuan uan
1717Integral lipatIntegral lipat
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Double Riemann Sums
In first year calculus, the definite integral was defined as a Riemann sum that gave
the area under a curve. There is a similar definition for the volume of a region
below a function of two variables. Let f(x,y) be a positive function of two variables
and consider the solid that is bounded below by f(x,y) and above a region R in the
xy-plane.
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For a two dimensional region, we approximated the area by adding up the areas of
many approximating rectangles. For the volume of a three dimensional solid, we
take a similar approach. Instead of rectangles, we use rectangular solids for the
approximation. We cut the region R into rectangles by drawing vertical and
horizontal lines in the xy-plane. Rectangles will be formed. We let the rectangles
be the base of the solid, while the height is the z-coordinate of the lower left
vertex. One such rectangular solid is shown in the figure.
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Dengan cara yang sama integral dapat juga dinytakan sebagai berikut :
Atau
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Taking the limit as the rectangle size approaches zero (and the number of rectangles
approaches infinity) will give the volume of the solid. If we fix a value of x and look at
the rectangular solids that contain this x, the union of the solids will be a solid with
constant width Dx. The face will be approximately equal to the area in the yz-plane
of the (one variable since x is held constant) function z = f(x,y).
This area is equal to
Untuk mendekati 0 integral menjadi :
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Volume
Cari volume yang dibatasi oleh g(x,y) dan fungsi
f (x,y) seperti pada gambar berikut. Click gambar
untuk melihat animasi
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Volume ( seperti soal sebelumnya )
Click gambar untuk melihat animasinya
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Volume Seperti soal sebelumnya
Click disini animation
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Solution
The cone is sketched below
We can see that the region R is the blue circle in the xy-plane. We can find the
equation by setting z = 0.
Solving for y (by moving the square root to the left hand side, squaring both sides, etc) gives
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Volume benda di Oktant pertama
Yang dibatasi oleh :
Click disini DPGraph Picture
Find the volume of the solid in the first octant
bounded by the graphs of http://www2.scc-fl.edu/lvosbury/images/Mpic132no32.gif
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Luas PermukaanCari luas permukaan yang ditentukan oleh
f(x,y) seperti pada gambar Click here untuk melihat animasi
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Fubini's Theorem
Let f, g1, g2, h1, and h2 be defined and continuous on a region R. Then the double
integral equals
Example
Set up the integral to find the volume of the solid that lies below the cone
and above the xy-plane. Solution
The cone is sketched below
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The "-" gives the lower limit and the "+" gives the upper limit. For the outer limits,
we can see that
-4 < x < 4
Putting this all together gives
Either by hand or by machine we can obtain the result
Volume = 64 /3
Notice that this agrees with the formula
Volume = r2h/3
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Example
Set up the double integral that gives the volume of the solid that lies below the sphere x2 + y2 + z2 = 6
and above the paraboloid z = x2 + y2 Solution
We substitute x2 + y2 + (x2 + y2)2 = 6or x2 + y2 + (x2 + y2)2 - 6 = 0
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Now factor with x2 + y2 as the variable to get (x2 + y2 - 2)(x2 + y2 + 3) = 0
The second factor has no solution, while the first is x2 + y2 = 2
Solving for y gives and - < x <
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Just as we did in one variable calculus, the volume between two surfaces is the
double integral of the top surface minus the bottom surface. We have
Again we can perform this integral by hand or by machine and get Volume = 7.74
Polar Double Integration FormulaBentuk integral dapat dibawa dalam bebntk koordinat polar terutama yang
berbentuk lingkatan yaitu x2 + y2. Untuk ini digunakan koordinat polar.
Sebagai contoh lihat berikut ini.
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Theorem: Double Integration in Polar Coordinates
Let f(x,y) be a continuous function defined over a region R bounded in polar
coordinates by r1() < r < r2() 1 < < 2
Then
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Even if r and are very small the area is not the product (r)(). This
comes from the definition of radians. An arc that extends radians a
distance r out from the origin has length r. If both r and are very
small then the polar rectangle has area
Area = r r
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The equation of the parabola becomes z = 9 - r2
We find the integral
This integral is a matter of routine. It evaluates to 28.
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Example
Find the volume of the part of the sphere of radius 3 that is left after
drilling a cylindrical hole of radius 2 through the center.
Solution
The picture is shown below
The region this time is the annulus (washer)
between the circles r = 2 and r = 3 as shown below.
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Example
Find the volume to the part of the paraboloid
z = 9 - x2 - y2
that lies inside the cylinder
x2 + y2 = 4
Solution
The surfaces are shown below.
The region R is the part of the xy-plane that is inside the cylinder. In polar coordinates, the
cylinder has equation r2 = 4
Taking square roots and recalling that r is positive gives
r = 2The inside of the cylinder is thus the polar
rectangle 0 < r < 2 0 < q < 2p
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The sphere has equation x2 + y2 + z2 = 9
In polar coordinates this reduces to
r2 + z2 = 9
Solving for z by subtracting r2 and taking a square root we
get top and bottom surfaces of
We get the double integral
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We get the double integral
This integral can be solved by letting u = 9 - r2 du = -2rdr
After substituting we get
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