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Please Discuss here about this GDB.Thanks

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MTH101 GDB No. 2 Starting Date: Thursday, Feb. 02, 2017 Closing Date: Thursday, Feb. 09, 2017 

Starting Date: Thursday, Feb. 02, 2017

Closing Date: Thursday, Feb. 09, 2017 at 11:59 P.M.


Ø  Post your solution of GDB only on GDB interface of LMS.

Ø  Do not post the solution of GDB on ‘regular MDB’ forum. It will not be graded on MDB.

Ø  There is preview option available on GDB forum. So, see the preview of your solution before posting your GDB solution.

Ø  Once GDB is posted, it cannot be edited or re-submitted.

Ø  In case, you see boxes instead of your solution in your GDB interface then email your GDB solution at mth101@vu.edu.pk before or on  due date.

      The following will be helpful to you for inserting mathematical equations in GDB interface:


Q. Find the volume of the solid generated when the region between the graphs of h(y)=3+y2   and w(y)=2−y2 over the interval [0,1]   is revolved about the y-axis.




M.Bilal Bhatti B$  thanks for sharing 


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As we know the farmula
v=ʃ(a to b) [π [{h(y)^2}-{w(y)^2}]dy
F(y)=h(y)=3+ y^2 and g(y)=w(y)=2- y^2
Then v will be as:
= ʃ(0 to 1) [π [{3+(y^2)^2}- {2-(y^2)^2}]dy
= ʃ(0 to 1) [π [(y^4+6y^2 +9)- (y^4- 4y^2)+4}]dy
= ʃ(0 to 1) [π (y^4+6y^2 +9- y^4+4y^2 -4)]dy
=ʃ(0 to 1) [ π (6y^2 +9+4y^2 -4)]dy
= ʃ(0 to 1) [π (10y^2 +5)]dy
= π (0 to 1) [{10ʃ(y^2) dy} + {5ʃ(5)dy}]
= π (0 to 1) [{(10 y^3)/3} +5y]
= π [{(10(1^3)/3) +5(1)} –{(10(0^3)/3)+5(0)}]
= π [{(10/3)+5} –{(0/3)+0}]
= π [{(10+15)/3} – {0}]
= π [{25/3}-0]
=π [25/3]
= [25/3] π

just copy and past

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you can use this to plot graphs

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slam to all

plz tell me how put Gdb 



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