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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.

BEST OF LUCK!

MTH 101 GDB SOLUTION

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M.Bilal Bhatti B$  thanks for sharing 

 

Note for All Members: You don’t need to go any other site for this assignment/GDB/Online Quiz solution, Because All discussed data of our members in this discussion are going from here to other sites. You can judge this at other sites yourself. So don’t waste your precious time with different links.

As we know the farmula
v=ʃ(a to b) [π [{h(y)^2}-{w(y)^2}]dy
Let,
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



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you can use this to plot graphs

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