wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the area of the region in the first quadrant enclosed by x-axis, the line y = 3x and the circle x2 + y2 = 16.

Open in App
Solution





x2 + y2 =16 represents a circle with centre O(0,0) and cutting the x axis at A(4,0)

y=3 x represents straight passing through O(0,0)

Point of intersection is obtained by solving the two equations

x2+y2 =16 and y=3 x ⇒x2+3 x2 =16⇒4x2 =16 ⇒x=±2⇒y =±23B2 ,23 and B'-2 ,-23 are points of intersection of circle and straight lineShaded areaOBQAO= areaOBPO+area PBQAP =∫023 x dx+∫2416-x2 dx =3x2202+12x16-x2+162sin-1x424 =23 +8×π2-23-8×π6 =23 +4π-23-4π3 =8π3 sq units Area bound by the circle and straight line above x axis =23 +-23+8×2π6 =8π3 sq units

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Sphere
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon