#### To determine

**a)**

**To estimate:**

The x coordinates of the points of intersection of the curves.

#### Answer

x=0 & x≈0.755

#### Explanation

**1) Concept:**

Draw the graph of the given curves, and find the point of intersection.

**2) Given:**

y=1-x2,y=x6-x+1

**3) Calculation:**

The equation of curve is

y=1-x2,y=x6-x+1

Draw the given curves.

From the graph, the given curves intersect at 0,1&0.755,0.430.

The x coordinates of the point of intersection are 0 & 0.755 .

**Conclusion:**

The x coordinates of point of intersection are 0 & 0.755.

#### To determine

**b)**

**To estimate:**

The area of R

#### Answer

≈0.12

#### 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:**

y=1-x2,y=x6-x+1

**3) Calculation:**

The equation of curve is

y=1-x2,y=x6-x+1

The points of intersection of the curves are 0,0&0.755,0.430.

From the graph for 0, 0.755, fx=1-x2≥gx=x6-x+1

Therefore, area becomes

A= ∫abfx-gx dx

=∫00.7551-x2-x6-x+1dx

=∫00.755x-x2-x6 dx

Integrating,

=x22-x33-x7700.755

Using fundamental theorem of calculus,

=(0.755)22-(0.755)33-(0.755)77-0

=0.12157919 ≈0.12

∫00.75x-x2-x6 dx≈0.12

**Conclusion:**

The area of R≈0.12

#### To determine

**c)**

**To estimate:**

The volume obtained by rotating R about x axis.

#### Answer

≈0.54

#### Explanation

**1) Concept:**

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:**

y=1-x2,y=x6-x+1

**3) Calculation:**

The equations of the curves intersect at 0,0& (0.755,0.43).

The cross section in plane Px has the shape of a washer with inner radius y=x6-x+1 and outer radius y=1-x2.

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

Ax=π(outer radius)2-π(inner radius)2

Ax=π(1-x2)2-π(x6-x+1)2

Ax=π(1-x2)2-(x6-x+1)2

Therefore,

V=∫00.755Axdx=π ∫00.755(1-x2)2-(x6-x+1)2dx

By simplifying the integrand,

π ∫00.75551-x22-x6-x+12dx

=π∫00.755[-x12+2x7-2x6+x4-3x2+2x]dx

By integrating,

V=π-x1313+2·x88-2·x77+x55-3·x33+2·x2200.755

Using fundamental theorem of calculus,

V=π-0.7551313+2·0.75588-2·0.75577+0.75555-3·0.75533+2·0.75522-π0

By simplifying,

V=0.544096 ≈0.54

V≈0.54

**Conclusion:**

The volume obtained by rotating R about x axis is ≈0.54.

#### To determine

**d)**

**To estimate:**

The volume obtained by rotating R about y axis.

#### Answer

≈0.31

#### 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 the 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 circumference, fx is height, and dx is the thickness of the shell.

**2) Given:**

y=1-x2,y=x6-x+1

**3) Calculation:**

Draw the curves y=1-x2& y=x6-x+1 about y axis.

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

It has radius =x.

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

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

fx=1-x2-(x6-x+1)

The curves y=1-x2,y=x6-x+1 intersect at 0,1&0.755,0.43.

So, a=0 and b=0.755

So the volume of the given solid is

V= ∫00.7552πx1-x2)-(x6-x+1 dx

V=2π∫00.755x·-x6-x2+xdx

V=2π∫00.755-x7-x3+x2dx

Integrating,

V=2π-x88-x44+x3300.755

Using fundamental theorem of calculus,

V=2π-0.75588-0.75544+0.75533-2π0

By simplifying,

V=308085 ≈0.31

V≈0.31

**Conclusion:**

The volume obtained by rotating R about y axis is ≈0.31.