#### To determine

**a)**

**To calculate:**

The area of R

#### Answer

512

#### Explanation

**1) Concept:**

The area A of the region bounded by the curves y=f(x), y=g(x) and the lines x=a and x=b, where f and g are continuous and fx≥gx for all x in a, b, is

A= ∫abfx-gxdx

**2) Given:**

The equations of the curvesare y=x3,y=2x-x2.

**3) Calculation:**

First find the intersection points of the curves by solving their equations simultaneously.

x3=2x-x2

x3+x2-2x=0

xx2+x-2=0

x(x+2)(x-1)=0

x=0, x=-2 and x=1

Since the region lies in the first quadrant, only x=0 and x=1 are the required values.

In 0, 1, fx=2x-x2≥gx=x3

So, the integration becomes

A= ∫abfx-gxdx

A= ∫01(2x-x2-x3)dx

Integrating,

A=2·x22-x33-x4401

Using fundamental theorem of calculus,

A=12-133-144-0

A=512

**Conclusion:**

The area of R is 512

#### To determine

**b)**

**To calculate:**

The volume obtained by rotating R about the x axis.

#### Answer

V=41105 π

#### Explanation

**1) Concept:**

If the cross section is a washer with the inner radius rin and outer radius rout, then the area of the washer is obtained by subtracting the area of the inner disk from the area of the outer disk.

A=π outer radius2 -π inner radius2

Let S be a solid that lies between x=a and x=b. If the cross sectional area of S in the plane Px, through x and perpendicular to the x-axis, is A(x), where A is continuous function, then the volume of S is

V=limn→∞∑i=1nAxi*∆x=∫abAxdx

**2) Given:**

The equations of the curves are, y=x3,y=2x-x2

**3) Calculation:**

Draw the curves y=x3& y=2x-x2 about x axis.

By drawing the graph of two curves, we find that the points of intersection are 0,0,1,1.

The cross sections are perpendicular to x axis.

The cross section in plane Px has the shape of a washer with the inner radius y=x3 and outer radius y=2x-x2.

So, find the cross sectional area by subtracting the area of the outer circle from the area of the inner circle, i.e.,

Ax=π(outer radius)2-π(inner radius)2

Ax=π2x-x22-πx32

Ax=π(2x-x2)2-x6

Therefore,

V=∫01Axdx

V=∫01π(2x-x2)2-x6dx

V=∫01π·[(4x2-4x3+x4)-(x6)] dx

Integrating,

V=π4·x33-4·x44+x55-x7701

Using fundamental theorem of calculus,

V=π4·133-14+155-177-π0

V=π43-1+15-17-0

V=41105π

**Conclusion:**

The volume obtained by rotating R about the x axis is 41105 π

#### To determine

**c)**

**To calculate:**

The volume obtained by rotating R about y axis.

#### Answer

V=1330 π

#### Explanation

**1) Concept:**

If x is the radius of a typical shell, then circumference=2πx and height is y.

By shell method, the volume of a solid by rotating the region under the curve y=f(x) about y- axis from a to b is

V= ∫ab2πxf(x)dx;a≤x≤b Where 2πx is the circumference, fx is height, and dx is the thickness of the shell.

**2) Given:**

The equations ofthe curves are y=x3,y=2x-x2

**3) Calculation:**

Draw the curves y=x3& y=2x-x2 about x axis.

The region and the typical shell are shown in the figure.

The graph shows the region and the height of cylindrical shell formed by rotation about the line y-axis

It has radius =x.

Using the shell method, find the typical approximating shell with radius x.

Therefore, the thickness is dx, circumference is 2πx, and height is

fx=2x-x2-x3

The curves y=x2,y=2x-x2 intersect at 0,0&1,1.

So, a=0 and b=1

So the volume of the given solid is

V= ∫012πx2x-x2-x3 dx

V=2π∫012x2-x3-x4dx

Integrating,

V=2π2·x33-x44-x5501

Using fundamental theorem of calculus,

V=2π2·133-144-155-2π0

V=2π23-14-15-0

V=1330 π

**Conclusion:**

The volume obtained by rotating R about the y axis is 1330 π