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Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Fig 9.46 - Mathematics

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प्रश्न

Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Fig 9.46

योग

उत्तर

By using integration we have to find the volume of the frustum.

So first find the equation of the curve.

Let A(0, 1) and B(2, 2) be two points.

Line joining these two points form a straight line.

That straight line revolves around x-axis.

Equation AB is `(y - 1)/(2 - 1) = (x  0)/(2 - 0)`

`(y - 1)/1 = x/2`

`y - 1 = x/2`

y = `x/2 + 1`

y = `(x + 2)/2`

Volume of the solid revolves around x-axis

= `pi int_"a"^"b" y^2  "d"x`

= `pi int_0^2 (x + 2)^2/4  "d"x`

= `pi/4 [(x + 2)^3/3]_0^2`

= `pi/4 [64/3 - 8/3]`

= `pi/4[56/3]`

= `14/3 pi  "m"^3` 

Volume of the frustum = `14/3 pi  "m"^3` 

shaalaa.com
Volume of a Solid Obtained by Revolving Area About an Axis
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Applications of Integration - Exercise 9.9 [पृष्ठ १३९]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 9 Applications of Integration
Exercise 9.9 | Q 5 | पृष्ठ १३९
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