Math 2443, Section 16.3

Transcription

1 Math 44, Section 6. Review These notes will supplement not replace) the lectures based on Section 6. Section 6. i) ouble integrals over general regions: We defined double integrals over rectangles in the last section. How does one define the double integral of a function f over a general region? Choose an arbitrary rectangle R which contains. efine a new function F on R as follows. F, y) f, y), on, outside. The function F is the same is f on and zero outside. Hence, it doesn t matter which rectangle we choose. If F is integrable over R, we can define, for a general region, f, y)da F, y)da. If the region is nice, the integral on the right-hand side will eist. ii) Type I and II regions: If a given region is of a specific type, it simplifies the double integral considerably. a) A planar region I is of Type I if it lies between the graphs of two continuous functions of. That is, I {, y) a b, g ) y g )} where g and g are continuous on [a, b]. R To find I f, y)da, we choose a rectangle R [a, b] [c, d] which contains I. Then, b d f, y)da F, y)da F, y)dyd I a c R Since F, y) for y g ) as well as y g ), and F, y) f, y) when g ) y g ). Thus, d c F, y)dy g ) g ) f, y)dyd. This gives us f, y)da I b g ) a g ) f, y)dyd.

2 Review b) A planar region II is of type II if it lies between the graphs of two continuous functions of y. That is, II {, y) c y d, h y) h y)} where h and h are continuous on [c, d]. Similar to above, we get II f, y)da d h y) c h y) f, y)ddy. Remark: Remember that in both the instances, the limits of the inner integral are functions of one variable but the integration is done with the opposite with respect to the other variable.. Eample: Evaluate the double integral cos y da, bounded by y, y,. Solution: We can write as {, y), y }. I have not drawn the region but you must do so. Clearly, is Type I region and thus cos yda cos y dyd siny] d sin d cos ] cos ). Eample: Evaluate the double integral y e y da, {, y) y 4, y}. Solution: The region is a Type II region, hence y y y e y da y e y d dy ye y y)dy ey y ] y y ey ] 4 dy e 6 7 ).. Eample: Find the volume of the solid under the surface z y and above the triangle with vertices, ), 4, ) and, ). Solution: The triangle with the given points as vertices can be seen to be a Type I region as follows. In the triangle, we see that lies between and 4 while

3 Math 44, Section 6. the y-component lies between y and the line joining the points, ) and 4, ), which has the equation y + 7. We thus write {, y) 4, y + 7 }. Again, I have not drawn the region but you have to do so. Note that the function f, y) y is always non-negative on. Hence, the volume V of the solid under the surface z y and above the given triangle is V y da +7 y y dyd ] +7 d ) ) + 7 d ) d ] ) 4 ) 64 ) + 6 ) 8..4 Eample: Find the volume of the tetrahedron bounded by the coordinate planes and the plane + y + z 6. Solution: The coordinate planes are given by, y and z. Let s find the vertices of the tetrahedron first. These will be the intersections of three of the four planes given. The intersection of, y and + y + z 6 is,, 6). Similarly, the other three vertices are,, ),,, ) and,, ). Now, the given tetrahedron is a solid that lies above the triangle in the y plane that has vertices, ),, ) and, ). Please draw this triangle and the tetrahedron, even though I have not done so. The line joining, ) and, ) is given by y +. We can write {, y), y + }, which is a Type I region. Now, the volume V of the tetrahedron is the double integral of the function 6 y over. V 6 y)da + 6y y y ] + d 6 ) + 6 y)dy d ) + ) ) + d ) d ) d 4 9 ]

4 4 Review.5 Eample: Find the volume of the solid enclosed by the cylinders z, y and the planes z, y 4. Solution: In these type of problems, it is not necessary to draw the given solid. You need to find the region in the y plane above which the solid is located. In this case, we know that the solid lies above the region bounded by the line y 4 and the parabolay in the y plane. In this region, lies between and. Hence {, y), y 4}, which is a Type I region. Hence, the volume of the solid is the double integral over of the function f, y). V da y ] 4 d ] dyd )d iii) Reversing the order of integration: For double integrals on general regions, we cannot simply switch the order of integration because the Fubini Theorem doesn t directly apply on such regions. Nonetheless, it is possible that a given double integral is difficult to solve and we d like to reverse the order of integration. For this, we first find the planar region on which the double integral is being evaluated. Net, if is originally a Type I region, we can switch the order of integration by epressing it as a Type II region and evaluating the integral..6 Eample: Evaluate the integral by reversing the order of integration. y e d dy. Solution: We can see that the inner integral cannot be solved by elementary methods. In order to reverse the order of integration we first find the region on which the integral is being evaluated. Looking at the limits on and y, we can epress this region as {, y) y, y}. This is a Type II region. It is given by the triangle in the y plane which is bounded by the lines y, y and. I have not drawn the region but you have to do so. Hence, y e d dy e da. Now, to switch the order of integration, we epress as a Type I region. To do this we must epress point, y) on such that lies between two real numbers

5 Math 44, Section 6. 5 and y is between two continuous functions of. It is clear that and also y. That is, Hence, {, y), y }. y e d dy 9 e da / ye ] / d eu du 6 e 9 ). where we have used the substitution u in the last step. e dyd e d.7 Eample: Evaluate the integral by reversing the order of integration. y + dyd. Solution: The double integral is taken over the region {, y) 4, y }, which is a Type I region and is represented by the region above the curve y, below the line y and between and 4. Now, we can epress this region as a Type II region as follows Therefore, {, y) y, y }. y + dyd 9 y + da ] y y + y dy u du ln9. y + ddy y y + dy where we have used the substitution u y + in the last step. iv) If a region is the union of two regions and. That is, then the double integral over is simply the sum of the double integrals over and. f, y)da f, y)da + f, y)da. Sometimes you won t be able to epress a given region as Type I or Type II. I such a case, see if you can write it as the union of two or more such regions.