#### To determine

**a)**

**To find:**

The volume of the solid rotating the region about x- axis

#### Answer

V=215 π

#### 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:**

The equations of the curves are y=x & y=x2.

**3) Calculation:**

The equations of curves are

y=x & y=x2

Draw the curves about x axis.

The curves y=x & y=x2 intersect at the points 0,0&1,1.

The region between them is the solid of rotation and across section perpendicular to x axis.

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

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=πx2-πx22

Ax=π(x)2-x4

Therefore,

V=∫01Axdx

V=∫01π(x)2-x4dx

Integrating,

V=πx33-x5501

Using fundamental theorem of calculus,

V=π13-15-π0

V=215 π

**Conclusion:**

V=215 π

#### To determine

**b)**

**To find:**

The volume of a solid rotating the region about y- axis

#### Answer

V=16 π

#### Explanation

**1) Concept:**

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

V=limn→∞∑i=1nAyi*∆y=∫abAydy

**2) Given:**

The equations of curves are y=x & y=x2.

**3) Calculation:**

The equations of curves are y=x & y=x2.

Draw the curve about y axis.

The curves x=y& x=y intersect at the points 0,0&1,1.

The region between them is the solid of rotation and across section perpendicular to y axis.

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

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

Ay=π(outer radius)2-π(inner radius)2

Ay=πy2-πy2

Ay=π(y)-y2

Therefore,

V=∫01Aydy

V=∫01π(y)-y2dy

Integrating,

V=πy22-y3301

V=π12-13-π0

V=16 π

**Conclusion:**

V=16 π

#### To determine

**c)**

**To find:**

The volume of the solid rotating the region about y=2.

#### Answer

V=815 π

#### 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:**

The equations of curves are y=x & y=x2.

**3) Calculation:**

The equations of curves are y=x & y=x2.

The line of revolution is y=2

The curves y=x & y=x2 intersect at the points 0,0&1,1.

The region between them is the solid of rotation and across section perpendicular to y=2 .

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

y=2-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=π2-x22-π2-x2

Ax=π2-x22-2-x2

Therefore,

V=∫01Axdx

V=∫01π2-x22-2-x2dx

V=∫01π·[(4-4x2+x4)-(4-4x+x2)]

By simplifying the integrand,

V=∫01π·x4-5x2+4xdx

Integrating,

V=πx55-5x33+4x2201

Using fundamental theorem of calculus,

V=π155-5·133+4·122-π0

V=π15-53+2-0

v=815 π

**Conclusion:**

v=815 π