Pertemuan

1. 17 Pertemuan Integral lipat Kalkulus II

2. 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.   Kalkulus II

3. 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.   Kalkulus II

4. Dengan cara yang sama integral dapat juga dinytakan sebagai berikut : Atau Kalkulus II

5. 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 : Kalkulus II

6. Volume Cari volume yang dibatasi oleh g(x,y) dan fungsi f (x,y) seperti pada gambar berikut. Click gambar untuk melihat animasi Kalkulus II

7. Volume ( seperti soal sebelumnya ) Click gambar untuk melihat animasinya Kalkulus II

8. Volume Seperti soal sebelumnya Click disini animation Kalkulus II

9. 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 Kalkulus II

10. 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 Kalkulus II

11. Luas Permukaan Cari luas permukaan yang ditentukan oleh f(x,y) seperti pada gambar Click here untuk melihat animasi Kalkulus II

12. Fubini's Theorem Let f, g1, g2, h1, and h2be 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 Kalkulus II

13. 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 p/3 Notice that this agrees with the formula          Volume  =  p r2h/3 Kalkulus II

14. 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  =  6 or   x2 + y2 + (x2 + y2)2 - 6  =  0 Kalkulus II

15. Now factor with x2 + y2as 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  <        Kalkulus II

16. 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 Formula Bentuk integral dapat dibawa dalam bebntk koordinat polar terutama yang berbentuk lingkatan yaitu x2 + y2.  Untuk ini digunakan koordinat polar. Sebagai contoh lihat berikut ini. Kalkulus II

17. Kalkulus II

18. Even if Dr and Dqare very small the area is not the product (Dr)(Dq).  This comes from the definition of radians.  An arc that extends Dq radians a distance r out from the origin has length rDq.  If both Dr and Dqare very small then the polar rectangle has area Area  =  r Dr Dq Kalkulus II

19. The equation of the parabola becomes     z  =  9 - r2 We find the integral  This integral is a matter of routine.  It evaluates to 28p. Kalkulus II

20. 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. Kalkulus II

21. 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  =  2 The inside of the cylinder is thus the polar rectangle          0  <  r  <  2        0  <  q  <  2p Kalkulus II

22. 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  Kalkulus II

23. We get the double integral  This integral can be solved by letting u  =  9 - r2           du  =  -2rdr After substituting we get Kalkulus II

24. Terima kasih Kalkulus II