#### To determine

**To describe:**

The solid for an integral π∫-111-y22dy which represents the volume of a solid

#### Answer

= x, y/-1≤y≤1, 0≤x≤1-y2 of the xy- plane about the y- axis

#### Explanation

**1) Concept:**

i. If the cross section is a disc and the radius of the disc is in terms of x or y then area A=π radius2

ii. The volume of solid revolution about y-axis is

V= ∫abA(y)dy

**2) Given:**

π∫-111-y22dy

**3) Calculation:** Compare the given integral with expression for volume.

So the cross section of the solid is a disc with radius 1-y2, and it is perpendicular to the axis of rotation y axis.

So, the solid is obtained by rotating the area bounded by the line x=0 and the curve x=1-y2, about the y-axis. And it lies between y= -1 and y=1

This can be written as

= x, y/-1≤y≤1, 0≤x≤1-y2

Therefore, the integral

π∫-111-y22dy

describes the volume of the solid obtained by rotating the region

= x, y/-1≤y≤1, 0≤x≤1-y2 of the xy- plane about the y- axis

**Conclusion:**

π∫-111-y22dy

above integral describes the volume of the solid obtained by rotating the region

= x, y/-1≤y≤1, 0≤x≤1-y2 of the xy- plane about the y- axis