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Introduction | Exercise 04[Volume of Solids]|

Volume of Solids

Volume by Double Integration

Consider a surface . Let the orthogonal projection on XY plane of its portion S’ be the surface S. Divide S into elementary rectangles of area  by lines parallel to x and y axis. Now consider prism have its base as these triangles and its length parallel to OZ.

The volume of this prism under consideration is bounded by surface S and surface  is .

Hence the volume of the solid cylinder having S as base, bounded by the given surfaces with generator parallel to Z-axis is

Where integration is to carried over surface S.

In polar coordinates the above integration becomes

Example:

Lets us find the volume bounded by the cylinder  and the planes  and .

Here, S’ i.e. is the plane , and its projection i.e. S on  will also be a plane surrounded by the base of the cylinder . Since the top plane is symmetrical about , so we first find half the volume and will double later to find actual volume.

Hence limits for the projected surface i.e.  are as: x varies from 0 to , and y varies from -2 to 2.

So, required Volume is

SEE ALSO

Volume of Solids as triple Integration, Volume of solids of revolutions, Change of variables in double integration, change of variables in triple integration

 

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