To create the cone, consider revolving a line through the origin about the $x$-axis.
The equation of this line is $y = \answer {\frac {3}{5} x}$.
We will integrate over the interval $\left (\answer {0},\answer {5}\right )$

The definite integral that gives the volume of the cone is:

The volume of the cone is: $V = \answer {15 \pi }$.

To create the cone, consider revolving a line through the origin about the $x$-axis.
The equation of this line is $y = \answer {\frac {R}{H} x}$.
We will integrate over the interval $\left (\answer {0},\answer {H}\right )$

The definite integral that gives the volume of the cone is:

The formula for the volume of a cone is: $V = \answer {\pi R^2 H/3}$.

The radius of the disk at $x$ is $r = \answer {2x+1}$
We will integrate over the interval $\left (\answer {0},\answer {2}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {\frac {62}{3} \pi }$.

To create the sphere, consider revolving a semi-circular region centered at the origin about the $x$-axis.
The equation of this semi-circle is $y = \answer {\sqrt {9-x^2}}$.
We will integrate over the interval $\left (\answer {-3},\answer {3}\right )$

The definite integral that gives the volume of the sphere is:

The volume of the sphere is: $V = \answer {36 \pi }$.

To create the sphere, consider revolving a semi-circular region centered at the origin about the $x$-axis.
The equation of this semi-circle is $y = \answer {\sqrt {R^2-x^2}}$.
We will integrate over the interval $\left (\answer {-R},\answer {R}\right )$

The definite integral that gives the volume of the sphere is:

The formula for the volume of a sphere with radius $R$ is: $V = \answer {4/3 \pi R^3 }$.

The radius of the disk at $x$ is $r = \answer {1-x^2}$
We will integrate over the interval $\left (\answer {-1},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {\frac {16}{15} \pi }$.

The radius of the disk at $x$ is $r = \answer {1-\sqrt x}$
We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {\frac 16 \pi }$.

The outer radius of the washer at $x$ is $R = \answer {1}$
The inner radius of the washer at $x$ is $r = \answer {1/x}$

We will integrate over the interval $\left (\answer {1},\answer {2}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {\pi /2 }$.

The outer radius of the washer at $x$ is $R = \answer {1}$
The inner radius of the washer at $x$ is $r = \answer {x^{1/3}}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {2\pi /5 }$.

The outer radius of the washer at $x$ is $R = \answer {1-x^2}$
The inner radius of the washer at $x$ is $r = \answer {1-x}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {\pi /5 }$.

The radius of the cylindrical shell at $x$ is $r = \answer {x}$
The height of the cylindrical shell at $x$ is $h = \answer {x+1}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {5\pi /3 }$.

The radius of the cylindrical shell at $x$ is $r = \answer {x}$
The height of the cylindrical shell at $x$ is $h = \answer {x-x^2}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {\pi /6 }$.

The radius of the cylindrical shell at $x$ is $r = \answer {x-1}$
The height of the cylindrical shell at $x$ is $h = \answer {1/x}$

We will integrate over the interval $\left (\answer {1},\answer {2}\right )$

The definite integral that gives the volume of the solid is:

The volume of the solid is $V = \answer {2\pi (1-\ln (2)) }$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$.
The radius and height as functions of $x$ are:
$r = \answer {x+1} \;\;\;$ and $\; h = \answer {\sqrt {x} - x}$.

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {7\pi /15}$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$.
The outer radius, $R$, and inner radius, $r$, as functions of $x$ are:
$R = \answer {1-x} \;\;\;$ and $\; r = \answer {1-\sqrt {x}}$.

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {\pi /6}$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {1}\;\;\;$ and $\;\;\; b = \answer {2}$.
The radius, $r$, as a function of $x$ is:
$r = \answer {2/x}.$

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {2\pi }$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$.
The radius, $r$, as a function of $x$ is:
$r = \answer {x^3}.$

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {\pi /7}$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {2}$.
The outer radius, $R$, and inner radius, $r$, as functions of $x$ are:
$R = \answer {4} \;\;\;$ and $\; r = \answer {4-x^2}$.

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {224\pi /15}$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {-1}\;\;\;$ and $\;\;\; b = \answer {\ln 2}$.
The radius, $r$, as a function of $x$ is:
$r = \answer {2e^x}.$

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {(8-2/{e^2})\pi }$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$.
The radius and height as functions of $x$ are:
$r = \answer {x} \;\;\;$ and $\; h = \answer {x^2 -x^3}$.

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {\pi /10}$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {3/2}$.
The outer radius, $R$, and inner radius, $r$, as functions of $x$ are:
$R = \answer {3-x} \;\;\;$ and $\; r = \answer {x}$.

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {27\pi /4}$.

The formula associated with this method is:

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {16}$.
The radius and height as functions of $x$ are:
$r = \answer {16-x} \;\;\;$ and $\; h = \answer {x^{1/4} }$.

The definite integral that represents the volume is given by:

The volume of the solid is $V = \answer {16384\pi /45}$.