    Question

# Area enclosed by curve y3−9y+x=0 and Y - axis is -

A
92
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
812
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
81
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is D 812The given equation of curve can be written as x=f(y)=9y−y3. Now to calculate the area we need to find the boundaries of this curve i.e ordinates or the point where this curve is meeting Y - axis. f(y)=y(9−y2) f(y)=y.(3+y)(3−y) So, the points where f(y) is meeting y - axis are y = -3, y = 0 & y = 3. Important thing to note here is that the function is changing its signs. I.e. from y = -3 to y = 0 f(y) is negative. & from y = 0 to y = 3 f(y) is positive. Since, the function in negative in the interval (-3, 0 ) we’ll take absolute value of it. Because we are interested in the area enclosed and not the algebraic sum of area. Let the area enclosed be A. A=∫0−3|f(y)|dy+∫30f(y)dyA=∫0−3|9y−y3|dy+∫30(9y−y3)dyA=−∫0−3(9y−y3)dy+∫309y−y3dyA=−[9y22−y44]0−3+[9y22−y44]30A=−[0−0−812+814]+[812−814]A=81−812A=812  Suggest Corrections  0      Similar questions  Related Videos   Property 6
MATHEMATICS
Watch in App  Explore more