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Find the area of the region bounded bythe Curvesy²= 2ax - x2 and y2°= ax in the first quadrant ? |
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Answer» The answerStep-by-step explanationSolution :x2+y2≤2axx2−2ax+y2+a2−a2≤0(x−a)2+y2≤a2y2≥ax(x−a)2+ax=a2x2+a2−2ax+ax=a2x(x−a)=0x=0,ay2=axy2=a⋅a=a2y=aArea of region=∫2a0ydx∫a0ax−−√dx+∫2aa2ax−x2−−−−−−−−√dxa−−√∫a0x−−√dx+∫2aaa2−(x−a)−−−−−−−−−−√dx23a2+[0+a2π4−0−0]=23a2+π4a2=a2(23+π4). |
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