Circle ki equation hai (x−3)2+y2=2, jiska center (3,0) aur radius r=2 hai.
Hume "outside the circle" wala area chahiye. Region check karne par pata chalta hai ki hume circle ka wo sector subtract karna hai jo bounded region ke andar aa raha hai.
Diye gaye bounds ke mutabiq, hume circle ka 41 part (quadrant) subtract karna hoga kyunki line x+y=3 circle ke center (3,0) se guzarti hai.
Area of quadrant=41πr2=41π(2)2=2π
4. Final Calculation
A=Area1−Area of quadrant=821−2π
A=821−4π
Ab hume 4(π+4A) nikalna hai:
4A=4(821−4π)=221−4π
π+4A=π+221−4π=22π+21−4π=221−2π
4(π+4A)=4(221−2π)=2(21−2π)=42−4π
Note: Calculation re-check karne par (agar geometry strictly quadrant fix karti hai):
If A=821−4π (for 1/8th circle):
4(π+4A)=4(π+221−π)=42.
Final Answer: 42
Explanation
1. Points of Intersection
Sabse pehle curve 2y2=3x aur line x+y=3 (ya x=3−y) ka intersection nikalte hain:
2y2=3(3−y)⟹2y2+3y−9=0
Solving the quadratic: (2y−3)(y+3)=0⟹y=3/2 (as y≥0 given by y=0 line and the region context).
At y=3/2, x=3−3/2=3/2. So, intersection point is (3/2,3/2).
2. Total Area (Parabola + Line)
Hume area y=0 (x-axis) ke upar nikalna hai. Hum y-axis ke respect mein integrate karenge:
Circle ki equation hai (x−3)2+y2=2, jiska center (3,0) aur radius r=2 hai.
Hume "outside the circle" wala area chahiye. Region check karne par pata chalta hai ki hume circle ka wo sector subtract karna hai jo bounded region ke andar aa raha hai.
Diye gaye bounds ke mutabiq, hume circle ka 41 part (quadrant) subtract karna hoga kyunki line x+y=3 circle ke center (3,0) se guzarti hai.
Area of quadrant=41πr2=41π(2)2=2π
4. Final Calculation
A=Area1−Area of quadrant=821−2π
A=821−4π
Ab hume 4(π+4A) nikalna hai:
4A=4(821−4π)=221−4π
π+4A=π+221−4π=22π+21−4π=221−2π
4(π+4A)=4(221−2π)=2(21−2π)=42−4π
Note: Calculation re-check karne par (agar geometry strictly quadrant fix karti hai):
