Let's start by sketching the enclosed region. We can find the points where the two curves intersect by setting the curves equal to each other.

So the two curves intersect at $x=0$‍ and at $x=2$‍, or the points $(0,0)$‍ and $(2,4)$‍.

If we rotate the enclosed region about the $x$‍-axis, the resulting solid will have a hole in the middle. So we can find the volume of the solid using the washer method.

The image below shows a washer cross section of the solid of revolution.

The volume of each washer can be found by multiplying the area of the face of each washer by the width of the washer, $dx$‍.

We can find the area of the face of each washer by subtracting the area of the inner ring from the area of the outer ring. Since the radius of the outer ring is $2x$‍ and the radius of the inner ring is $x^2$‍, we get:

So the volume of each washer is $v(x)=\pi (4x^2-x^4)dx$‍.

To find the volume of the entire solid, we sum up the volumes of infinitely many of these washers between $x=0$‍ and $x=2$‍, and so the volume of the entire solid is given by $V(x)={\displaystyle {\int }_0^2\pi (4x^2-x^4)dx}$‍.

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{64\pi }{15}}$‍cubic units.

Let's start by sketching the enclosed region. Remember, we can find the points where the two curves intersect by setting the curves equal to each other.

In this problem, the enclosed region is not rotated about the $x$‍-axis, but instead about a line parallel to the $x$‍-axis, $y=-1$‍.

The image below shows a washer cross section of the solid of revolution.

Since the region is rotated about the line $y=-1$‍, the radius of the outer ring is $\sqrt{x}+1$‍ and the radius of the inner ring is $x^2+1$‍. We can find the area of the face of each washer by subtracting the area of the inner ring from the area of the outer ring.

The volume of each washer can now be found by multiplying the area of the face of each washer by the width of the washer, $dx$‍, and so $v(x)=\pi (x+2\sqrt{x}-x^4-2x^2)dx$‍.

To find the volume of the entire solid, we sum up the volumes of infinitely many of these washers between $x=0$‍ and $x=1$‍, and so the volume of the entire solid is given by $V(x)={\displaystyle {\int }_0^1\pi (x+2\sqrt{x}-x^4-2x^2)dx}$‍.

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{29\pi }{30}}$‍cubic units.

Let's start by sketching the enclosed region.

In this problem, note that the enclosed region is rotated about the $y$‍-axis not the $x$‍-axis.

This creates horizontal washer cross sections as shown below.

Notice that the washers are horizontal, and so the width of each washer is $dy$‍. Therefore, we should find all of our components in terms of $y$‍.

If we write each equation as a function of $y$‍, we see that the radius of the outer ring is $y$‍ and the radius of the inner ring is $y^2$‍.

We can find the area of the face of each washer by subtracting the area of the inner ring from the area of the outer ring.

The volume of each washer can now be found by multiplying the area of the face of each washer by the width of the washer, $dy$‍, and so $v(y)=\pi (y^2-y^4)dy$‍.

To find the volume of the entire solid, we sum up the volumes of infinitely many of these washers between $y=0$‍ and $y=1$‍, and so the volume of the entire solid is given by $V(y)={\displaystyle {\int }_0^1\pi (y^2-y^4)dy}$‍.

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{2\pi }{15}}$‍cubic units.

Let's start by sketching the enclosed region.

In this problem, the enclosed region is not rotated about the $y$‍-axis, but instead about a line parallel to the $y$‍-axis, $x=4$‍.

Let's sketch this region and a cross section of the solid.

Notice that the washers are horizontal, and so the width of each washer is $dy$‍. Therefore, we should find all of our components in terms of $y$‍.

If we write each equation as a function of $y$‍, we see that the radius of the outer ring is $4-{\displaystyle \frac{1}{4}}{y}^2$‍ and the radius of the inner ring is $4-y$‍.

We can find the area of the face of each washer by subtracting the area of the inner ring from the area of the outer ring.

The volume of each washer can now be found by multiplying the area of the face of each washer by the width of the washer, $dy$‍, and so $v(y)=\pi ({\displaystyle \frac{1}{16}}{y}^4-3y^2+8y)dy$‍.

To find the volume of the entire solid, we sum up the volumes of infinitely many of these washers between $y=0$‍ and $y=4$‍, and so the volume of the entire solid is given by $V(y)={\displaystyle {\int }_0^4\pi ({\displaystyle \frac{1}{16}}{y}^4-3y^2+8y)dy}$‍.

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{64\pi }{5}}$‍cubic units.