find the area enclosed by the curve x=3cost, y=2sint Share with your friends Share 2 Manisha Sachdeva answered this Dear student, x=3 cos t and y=2 sin t2x = 6 cos t ....1 and 3y =6 sin t ....2Squaring and adding 1 and 2, we get 4x2+9y2=36⇒x29 + y24 = 1This is the equation of the ellipse.Now we find the area enclosed by the ellipseArea=4∫03239-x2dx =83∫039-x2dx =83x29-x2+92sin-1x303 =83329-9+92sin-133-029+92sin-103 =83×9π4 =6π sq units Regards 12 View Full Answer Pradeep Ch answered this Hint: x=3cost 2x=6cost y=2sint 3y=6sint Clearly, (2x)2+(3y)2 = 36(sin2t+cos2t) = 36 so, 4x2+9y2=36, which is the equation of an ellipse. Find area by integration for one quadrant and multiply it by 4. 4
