#### To determine

**To find:**

Integral for the volume of cut out

#### Answer

V=8∫0rr2-y2·R2-y2dy

#### Explanation

**1) Concept:**

i) Solid of Revolution:

Volume of solid revolution revolving a region around a line is given by

V=∫abA(x)dx or V=∫cdA(y)dy

In case of the disk method, the radius is found in terms of x or y and use A=π·radius2

In case of the washer method, find the inner and outer radius and compute the area of washer by subtracting the area of inner disk from outer disk, use

A=π·outer radius2-π·inner radius2

ii) Integrals of Symmetric functions:

Suppose f is continuous on [-a, a]

a) If f is even [f-x=fx], then ∫-aaf(x)dx=2∫0af(x)dx

b) If f is odd [f-x=-fx], then ∫-aaf(x)dx=0

**2) Calculation:**

Now from the front side, the circle appears to be in a y-z plane,

Therefore, equation of circle is y2+z2=R2

From the top side, the hole lies on a x-y plane,

Therefore, equation of circle is x2+y2=r2

From the front side, the diagram will appear as shown below:

Here, the x- axis is the axis of cylindrical hole of radius r. A quarter of the cross section through y, perpendicular to the y- axis, is the rectangle shown.

By using Pythagoras theorem,

x=r2-y2 and z=R2-y2

So,

14·Ay=xz

Ay=4r2-y2·R2-y2

V=∫abA(y) dy

Since the cut out varies from –r to r

V=∫-rr4r2-y2·R2-y2dy

V=4·∫-rrr2-y2·R2-y2dy

By using concept of integrals of symmetric function,

V=8∫0rr2-y2·R2-y2dy

So, the above value obtained is the volume of thecut out portion.

**Conclusion:**

V=8∫0rr2-y2·R2-y2dy