Lecture Wise Questions from 23 to 45 By Virtualians.pk. Q105. What is the impact of double integration in finding out the area and volume of Regions?

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1 Lecture Wise Questions from 23 to 45 By Virtualians.pk Q105. What is the impact of double integration in finding out the area and volume of Regions? Ans: It has very important contribution in finding the areas and volume of the certain regions and objects. And as we did in our examples we have the function and the domain of that function and simply calculate the volume of 3d object which is formed by the surface of the function and the region on which our surface is defined and also double integral are useful for finding out the region in the plane. Q106. How can we take polar co ordinates of a circle? Ans: You can get the polar equation of the circle rather then the coordinate of the circle because circle is the locus of points whose distance from a fixed point remains the same. And by joining these points you get the circle and remember that you always talk about the coordinates of a point. And circle is not a point; circle is the set of points Now we did in our lecture that if you have a circle x + y = a in rectangular coordinates with center (0, 0) and radius "a" then polar equation of that circle is r = a. Which is obtained by using the relation between the rectangular and polar coordinates which is given by the following y x = r cos θ, y = r sin θ, r = x + y and θ = tan equations x. Q107. How to find out the limits of Double integrals in Polar coordinates? Ans: Well First of all you will draw the region then you can get the limits for "r" simply by drawing a line from Pole with some constant angle"θ " It will intersect your region of integration now at two points. And the point where the line enters the region is your lower limit and the point of intersection where the line leaves your region is the Upper limit for "r". Now for the limit of "theta" you will rotate a radial line in anti clock wise direction and you will note the angle at which this line enters your region, this will be your lower limit for "Theta" and where it leaves the region you will again note the angle of that radial line, which will be the upper limit of your double integral in Polar coordinates. For examples you should see the Lecture # 25 and i hope that you will under these examples now. Prepared by:irfan Khan Page 1

2 Q108. How can we change Cartesian integrals into Polar integrals? Ans: If you have a double integral in Cartesian coordinates then you can convert this integral into Polar coordinates now as you know that the relation between Cartesian coordinates and Polar coordinates is x= rcos θ, y = rsinθ replace "x" and "y" by this in the integral. Also you know the region of integration and you have to convert the boundary curves in polar coordinates and then apply the method of obtaining limits in polar coordinates. Then your Double integral will convert to the Polar coordinates. Also note that if you have equation of circle in rectangular coordinates and we convert the rectangular equation of the same circle into Polar coordinates but if you draw that equation in polar coordinates it will be again the same circle which we have in rectangular coordinates and we also did in our lectures. Now what is the conclusion, well conclusion is that your region of integration remain the same if you change the double integral from one coordinate system to the other only limits will change and this is because the equation of the same curves which form the region of integration is different in different coordinates axes as clear from the answer of above question. Q109. Explain the second example of lecture # 26 how we obtain the limits of the integral in detail? 2 2 x + y e da Ans: In this example we have to find out the double integral R where our 2 region of integration is formed by semi circle whose equation is given by y = 1 x and the x-axis. Now you want to know how we get the limits of the double integral when we convert this integral from rectangular coordinates to the Polar coordinates. Well you can draw the 2 region of integration very easily. As y = 1 x represent the upper part of the circle with centre at (0, 0) and radius 1. Now for the limit of "r" you draw a line from pole that will intersect your region of integration at two points, now by drawing that line you can easily see that this line enters the region at r=0 which is the lower limit of the double integral and it leaves the region through the polar curve r=1 which is upper limit for "r". Remember that the equation of the circle with centre at (0, 0) and radius "a" in polar coordinates is r =a. So in particular equation of circle with center at (0, 0) and radius 1 in Polar coordinates is r =1. Prepared by:irfan Khan Page 2

3 Now for the limits of "θ " rotate a line through pole in anticlockwise direction and note that the angle at which that line enters the region and when it leaves the region and you will note that it enters the region atθ = 0 which is lower limit of the "θ " and leave the region at the angle at θ = π which is the upper limit for " θ " as shown in the figure below. Prepared by:irfan Khan Page 3

4 I hope that you will get the concept how we get these limits. Q110. What is the proper definition of vector valued functions? Ans: As you know that function from a set to the other set is rule which takes every element of one set to the other set. Now suppose that we have a function f : A B where A is the set of real numbers and B is the set of vectors in 3d or 2d. It means that every element of A (which is a real number) is map onto a vector in 3d. In simple language we can say that image of each real number of the set A is a vector in the set B. Such a function is known as vector valued function. Q111. What do we mean by the parametric equations of a curve? Ans: Suppose that you have a graph which does not satisfy the vertical line test. Remember that the vertical line test is used to get the information about the graph that whether it is graph of a function or not and you have learnt this test in your first course of calculus. Consider any arbitrary point P(x, y) on this curve now we try to represent the coordinate of this point namely "x" and "y" as function of another variable "t" say which is known as the parameter for this graph or curve. You can say that parameter is also a variable which is used to define the coordinates of arbitrary point of a graph which is not graph of a Prepared by:irfan Khan Page 4

5 function. In such a case we will define both x and y coordinates of arbitrary point on the curve in terms of another variable "t" and the expressions of x and y are functions. For example the parametric equations for the circle of radius "a" are x= acos θ, y = asinθ here "θ " is a parameter. Q112. What are line integrals? Ans: We have a function defined along a curve; now remember that function may be Scalar function or a vector valued function. We can define the integral along that curve which is known as line integral of that function along the curve. So line integral is the integration of a function along a curve. Now as you know that for a curve we will divide it into small arc lengths along which we will integrate the function. That s why in line integral we have "ds" which represent a small change in the arc length. Q113. What is Alternative form of a line integral? Ans: As you know that we define the line integral along a curve and in terms of arc length. We can also get the same line integral in the form of "dx" and "dy" which is known as the alternative form of the Line integral. Now how these two definitions are equivalent. Well we define the line integral of Fds function "F" say along the curve "C" then it is C where ds is the arc length of the curve along which we are going to integrate the function F. Now if we have components of the function F as P and Q in the direction of x and y as shown in the figure below. Prepared by:irfan Khan Page 5

6 Then we can write this line integral as form of line integral. C ( Pdx + Qdy) Q114. What the purpose of "integration of exact differentials? which is known as the alternative Ans: Differential equations are much more important in our real life. Because many problems of the real life when converted into mathematical model then we got the Differential equation. So if we want to find out the solution of that problem of the real life problems we will look for the solution of those Differential equations. Exact differential equations are the special differential equation and we solve these using Integration. Lecture 31 to 45 Lecture wise Questions Question: what are vector valued functions? Answer: The functions for which the domain consists of real numbers and the range consist of vectors in space or plane, such functions are called vector valued functions. Example: For x=2 f(x)=2x 2 i + x 3 j +3k k f(2)=8 i + 8 j +3k k Question: Explain line integral of function with respect to arc length. Answer: The line integral of along C is denoted by, We use a ds here, that we are moving along the curve, C, instead of the x-axis (denoted by dx) or the y-axis (denoted by dy). Because of the ds we call the line integral of f with respect to arc length. b 2 2 dx dy L = ds, where ds = + dt dt dt a Prepared by:irfan Khan Page 6

7 The line integral of f with respect to x is, C b f ( x, y) dx = f ( x(), t y()) t x () t dt a while the line integral of f with respect to y is, C b f ( x, y) dy = f ( x(), t y()) t y () t dt a Note that the only notational difference between these two and the line integral with respect to arc length (from the previous section) is the differential. These have a dx or dy while the line integral with respect to arc length has a ds. So when evaluating line integrals be careful to first note which differential you ve got so you don t work the wrong kind of line integral. These two integral often appear together and so we have the following shorthand notation for these cases. Question: can two tangents be drawn from a point on a curve? Answer: 1 and only 1 tangent can be drawn from a single point on a curve. Question: What is a sphere? Answer: sphere is the set of all points in space that are equidistant from a fix point in space Example: x 2 + y 2 + z 2 =4 is the set of all points in space whose distance from a fix point i.e. center is same (2). Prepared by:irfan Khan Page 7

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