Math, asked by kguddu0000, 1 year ago

Find the area common to the circle x2 +
y2 = 25 and parabola 3x2 = 16y.

Answers

Answered by kandsarveshpk

Answer:

The required area is area of the circle x2+y2=16 which is extension to the parabola y2=6x

To find the points of intersection,let us solve the given equation.

substituing for y2 in the equation of the circle we get,

x2+6x=16

⇒x2+6x−16=0

on factorising we get ,

(x+8)(x−2)=0

⇒x=−8andx=2.

Step 2:

since for x=-8 we get y2=−8 which is imaginary,we cannot take this value.

Now if x=2 we get y=±23–√.

Hence the points of intersection are (2,23–√) and (2,−23–√).

Now let us take the limits as 0 to 2 for parabola and 2 to 4 for the circle.

The area bounded by the circle and parabola is 2x[area bounded by the parabola and x-axis +area bounded by the circle and x-axis] A=2×[∫206x−−√dx+∫4216−x2−−−−−−√dx]

on integrating we get

A=2[6–√x3232]20+x2[16−x2−−−−−−√+162sin−1x4]42

Step 3:

on applying limits we get,

A=46–√3(22–√)+2[0+162sin1(1)−16−4−−−−−√+162sin−112]

A=46–√3(22–√)+2[8π2−43–√+8(π6)]

on simplifying we get,

A=43[43–√+6π−33–√−2π]

=43[3–√+4π]

Area of the circle=πr2

⇒π(4)2

⇒16π sq.units.

The required area =16π−43[4π+3–√]

⇒43[4×3π−4π−3–√]

⇒43[8π−3–√]sq.units.

Hence the required area=43[8π−3–√]sq.units.

Answered by priyansh9397