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Application Of Integrals

Question
CBSEENMA12035754
Wired Faculty App

Using integration, find the area of the region enclosed between the two circles:
straight x squared plus straight y squared space equals space 4 space and space left parenthesis straight x minus 2 right parenthesis squared plus straight y squared space equals space 4.

Solution

Given equations of the circles are:
straight x squared plus straight y squared space equals space 4 space space space... left parenthesis 1 right parenthesis left parenthesis straight x minus 2 right parenthesis squared plus straight y squared space equals space 4 space... left parenthesis 2 right parenthesis
Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle with centre C(2, 0) and radius 2.
Solving (1) and (2), we have:
left parenthesis straight x minus 2 right parenthesis squared plus straight y squared space equals space straight x squared plus straight y squared straight x squared minus 4 straight x plus 4 plus straight y squared space equals space straight x squared plus straight y squared straight x space equals space 1
This gives straight y equals plus-or-minus square root of 3
Thus, the points of intersection of the given circles are straight A open parentheses 1 comma space square root of 3 close parentheses space and space straight A apostrophe left parenthesis 1 comma space minus square root of 3 right parenthesis space as space shown space in space the space figure. space

Required area
 = Area of the region OACA'O
 = 2[area of the region ODCAO] 
 =2[area of the region ODAO + area of the region DCAD]
equals 2 open square brackets integral subscript 0 superscript 1 ydx plus integral subscript 1 superscript 2 ydx close square brackets equals 2 open square brackets integral subscript 0 superscript 1 square root of 4 minus left parenthesis straight x minus 2 right parenthesis squared end root dx plus integral subscript 1 superscript 2 square root of 4 minus straight x squared end root dx close square brackets equals 2 open square brackets 1 half left parenthesis straight x minus 2 right parenthesis square root of 4 minus left parenthesis straight x minus 2 right parenthesis squared end root plus 1 half cross times 4 sin to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator 2 end fraction close parentheses close square brackets subscript 0 superscript 1 plus 2 open square brackets 1 half straight x square root of 4 minus straight x squared end root plus 1 half cross times 4 sin to the power of negative 1 end exponent straight x over 2 close square brackets subscript 1 superscript 2 equals open square brackets left parenthesis straight x minus 2 right parenthesis square root of 4 minus left parenthesis straight x minus 2 right parenthesis squared end root plus 4 sin to the power of negative 1 end exponent open parentheses fraction numerator straight x minus 2 over denominator 2 end fraction close parentheses close square brackets subscript 0 superscript 1 plus open square brackets straight x square root of 4 minus straight x squared end root plus 4 sin to the power of negative 1 end exponent straight x over 2 close square brackets subscript 1 superscript 2 equals open square brackets negative square root of 3 plus 4 sin to the power of negative 1 end exponent open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses minus 4 sin to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis close square brackets plus open square brackets 4 sin to the power of negative 1 end exponent 1 minus square root of 3 minus 4 sin to the power of negative 1 end exponent 1 half close square brackets equals open square brackets open parentheses negative square root of 3 minus 4 cross times straight pi over 6 close parentheses plus 4 cross times straight pi over 2 close square brackets plus open square brackets 4 cross times straight pi over 2 minus square root of 3 minus 4 cross times straight pi over 6 close square brackets equals fraction numerator 8 straight pi over denominator 3 end fraction minus 2 square root of 3

Some More Questions From Application of Integrals Chapter

A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ` 100 and that on a bracelet is ` 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced.

Solve the following L.P.P. graphically :
Minimise Z = 5x + 10y
Subject to x + 2y ≤ 120
Constraints x + y ≥ 60
x – 2y ≥ 0
and x, y ≥ 0

The line L1: y = x = 0 and L2: 2x + y = 0 intersect the line L3: y + 2 = 0 at P and Q respectively. The bisectorof the acute angle between L1 and L2 intersects L3 at R.
Statement-1: The ratio PR: RQ equals 2√2:√5
Statement-2: In any triangle, the bisector of an angle divides the triangle into two similar triangles.

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