wiz-icon
MyQuestionIcon
MyQuestionIcon
6
You visited us 6 times! Enjoying our articles? Unlock Full Access!
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 C 812
The given equation of curve can be written as x=f(y)=9yy3. 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(9y2)
f(y)=y.(3+y)(3y)
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=03|f(y)|dy+30f(y)dyA=03|9yy3|dy+30(9yy3)dyA=03(9yy3)dy+309yy3dyA=[9y22y44]03+[9y22y44]30A=[00812+814]+[812814]A=81812A=812

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 6
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon