Answer
Verified
430.5k+ views
Hint: Here, we will draw the R region of Integration only by using the limits of Integration and not by evaluating the Integral. Then by using the R region of Integration, we will find the order of integration after reversing. Integration is defined as the summation of all the discrete data.
Complete Step by Step Solution:
We are given an Integral function \[\int_0^1 {\int_{ - \sqrt {1 - {y^2}} }^{\sqrt {1 - {y^2}} } {dxdy} } \].
The given Integration limit shows that the region is a unit circle, which gives the underlying relationships as:
\[{x^2} + {y^2} = 1\]
Now, the limits of the Integration of the variables \[x\] and \[y\] are
\[ \Rightarrow x = \pm \sqrt {1 - {y^2}} \] and \[y = \pm \sqrt {1 - {x^2}} \]
But, the given limits of the Integration are:
\[ - \sqrt {1 - {y^2}} \le x \le \sqrt {1 - {y^2}} \] and \[0 \le y \le 1\]
We will draw the R Region of Integration for the given limits in the first Quadrant \[{Q_1}\] and the second Quadrant \[{Q_2}\].
Now, we will change the order of Integration.
\[ - 1 \le x \le 1\] and \[0 \le y \le \sqrt {1 - {x^2}} \]
Thus, we will get the Integration after reversing the order of integration as \[\int_{ - 1}^1 {\int_0^{\sqrt {1 - {x^2}} } {dydx} } \].
Therefore, the Integration after reversing the order of integration is \[\int_0^1 {\int_{ - \sqrt {1 - {y^2}} }^{\sqrt {1 - {y^2}} } {dxdy} } = \int_{ - 1}^1 {\int_0^{\sqrt {1 - {x^2}} } {dydx} } \].
Note:
We know that R region is a region that is common for the given functions and lies in the quadrants. The order of Integration is a process of reversing the order from \[dxdy\] to\[dydx\]. \[\int_a^b {f\left( x \right)dx} \] where \[a\] is the lower limit of the Integral and \[b\] is the upper limit of the Integral. We know that Integration is the process of adding the small parts to find the whole parts. Double Integration is defined as the process of Integration over a two Dimensional area.
Complete Step by Step Solution:
We are given an Integral function \[\int_0^1 {\int_{ - \sqrt {1 - {y^2}} }^{\sqrt {1 - {y^2}} } {dxdy} } \].
The given Integration limit shows that the region is a unit circle, which gives the underlying relationships as:
\[{x^2} + {y^2} = 1\]
Now, the limits of the Integration of the variables \[x\] and \[y\] are
\[ \Rightarrow x = \pm \sqrt {1 - {y^2}} \] and \[y = \pm \sqrt {1 - {x^2}} \]
But, the given limits of the Integration are:
\[ - \sqrt {1 - {y^2}} \le x \le \sqrt {1 - {y^2}} \] and \[0 \le y \le 1\]
We will draw the R Region of Integration for the given limits in the first Quadrant \[{Q_1}\] and the second Quadrant \[{Q_2}\].
Now, we will change the order of Integration.
\[ - 1 \le x \le 1\] and \[0 \le y \le \sqrt {1 - {x^2}} \]
Thus, we will get the Integration after reversing the order of integration as \[\int_{ - 1}^1 {\int_0^{\sqrt {1 - {x^2}} } {dydx} } \].
Therefore, the Integration after reversing the order of integration is \[\int_0^1 {\int_{ - \sqrt {1 - {y^2}} }^{\sqrt {1 - {y^2}} } {dxdy} } = \int_{ - 1}^1 {\int_0^{\sqrt {1 - {x^2}} } {dydx} } \].
Note:
We know that R region is a region that is common for the given functions and lies in the quadrants. The order of Integration is a process of reversing the order from \[dxdy\] to\[dydx\]. \[\int_a^b {f\left( x \right)dx} \] where \[a\] is the lower limit of the Integral and \[b\] is the upper limit of the Integral. We know that Integration is the process of adding the small parts to find the whole parts. Double Integration is defined as the process of Integration over a two Dimensional area.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE