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