Use the volume of revolution formula about the x-axis:

\(V=\pi\int y^2\,dx\).

Here \(y=x\), so

\(V=\pi\int_0^2 x^2\,dx=\pi\left[\dfrac{x^3}{3}\right]_0^2=\dfrac{8\pi}{3}\).

Use \(V=\pi\int y^2\,dx\).

Since \(y=x^2\),

\(V=\pi\int_0^1 (x^2)^2\,dx=\pi\int_0^1 x^4\,dx\).

\(V=\pi\left[\dfrac{x^5}{5}\right]_0^1=\dfrac{\pi}{5}\).

Use \(V=\pi\int y^2\,dx\).

Since \(y=3x\),

\(V=\pi\int_0^2 (3x)^2\,dx=9\pi\int_0^2 x^2\,dx\).

\(V=9\pi\left[\dfrac{x^3}{3}\right]_0^2=9\pi\cdot\dfrac{8}{3}=24\pi\).

Use \(V=\pi\int y^2\,dx\).

So

\(V=\pi\int_0^2 (2-x)^2\,dx\).

Expand: \((2-x)^2=x^2-4x+4\).

Then

\(V=\pi\left[\dfrac{x^3}{3}-2x^2+4x\right]_0^2=\pi\left(\dfrac{8}{3}\right)=\dfrac{8\pi}{3}\).

Use \(V=\pi\int y^2\,dx\).

Since \(y=\sqrt{x}\), we have \(y^2=x\).

So

\(V=\pi\int_0^4 x\,dx=\pi\left[\dfrac{x^2}{2}\right]_0^4=8\pi\).

Use \(V=\pi\int y^2\,dx\).

Here \(y=x^{3/2}\), so \(y^2=x^3\).

Hence

\(V=\pi\int_0^1 x^3\,dx=\pi\left[\dfrac{x^4}{4}\right]_0^1=\dfrac{\pi}{4}\).

Use \(V=\pi\int y^2\,dx\).

Since \(y=\dfrac{1}{x}\), we have \(y^2=\dfrac{1}{x^2}=x^{-2}\).

So

\(V=\pi\int_1^2 x^{-2}\,dx=\pi\left[-x^{-1}\right]_1^2\).

\(V=\pi\left(-\dfrac12-(-1)\right)=\dfrac{\pi}{2}\).

Use \(V=\pi\int y^2\,dx\).

Here

\((x^2+1)^2=x^4+2x^2+1\).

So

\(V=\pi\int_0^1 (x^4+2x^2+1)\,dx\)

\(=\pi\left[\dfrac{x^5}{5}+\dfrac{2x^3}{3}+x\right]_0^1\)

\(=\pi\left(\dfrac15+\dfrac23+1\right)=\dfrac{28\pi}{15}\).

Use \(V=\pi\int y^2\,dx\).

Since \(y=2x\),

\(V=\pi\int_0^3 (2x)^2\,dx=4\pi\int_0^3 x^2\,dx\).

So

\(V=4\pi\left[\dfrac{x^3}{3}\right]_0^3=4\pi\cdot9=36\pi\).

Use \(V=\pi\int y^2\,dx\).

So

\(V=\pi\int_0^2 (4-x^2)^2\,dx\).

Expand:

\((4-x^2)^2=16-8x^2+x^4\).

Then

\(V=\pi\left[16x-\dfrac{8x^3}{3}+\dfrac{x^5}{5}\right]_0^2\)

\(=\pi\left(32-\dfrac{64}{3}+\dfrac{32}{5}\right)=\dfrac{256\pi}{15}\).

Use \(V=\pi\int y^2\,dx\).

\((x+1)^2=x^2+2x+1\).

So

\(V=\pi\int_0^2 (x^2+2x+1)\,dx\)

\(=\pi\left[\dfrac{x^3}{3}+x^2+x\right]_0^2\)

\(=\pi\left(\dfrac{8}{3}+4+2\right)=\dfrac{26\pi}{3}\).

Use \(V=\pi\int y^2\,dx\).

First expand \(y\):

\(y=2x-x^2\).

So

\(y^2=(2x-x^2)^2=4x^2-4x^3+x^4\).

Then

\(V=\pi\int_0^2 (4x^2-4x^3+x^4)\,dx\)

\(=\pi\left[\dfrac{4x^3}{3}-x^4+\dfrac{x^5}{5}\right]_0^2\)

\(=\pi\left(\dfrac{32}{3}-16+\dfrac{32}{5}\right)=\dfrac{16\pi}{15}\).

Use \(V=\pi\int y^2\,dx\).

Since \(y=x^3\), we have \(y^2=x^6\).

So

\(V=\pi\int_0^1 x^6\,dx=\pi\left[\dfrac{x^7}{7}\right]_0^1=\dfrac{\pi}{7}\).

Use \(V=\pi\int y^2\,dx\).

Expand:

\((2\sqrt{x}+1)^2=4x+4\sqrt{x}+1\).

So

\(V=\pi\int_0^1 (4x+4x^{1/2}+1)\,dx\)

\(=\pi\left[2x^2+\dfrac{8}{3}x^{3/2}+x\right]_0^1\)

\(=\pi\left(2+\dfrac{8}{3}+1\right)=\dfrac{17\pi}{3}\).

Use \(V=\pi\int y^2\,dx\).

So

\(V=\pi\int_1^3 (3-x)^2\,dx\).

Expand:

\((3-x)^2=x^2-6x+9\).

Then

\(V=\pi\left[\dfrac{x^3}{3}-3x^2+9x\right]_1^3\)

\(=\pi\left(9-\dfrac{19}{3}\right)=\dfrac{8\pi}{3}\).

For rotation about the y-axis, use

\(V=\pi\int x^2\,dy\).

Since \(x=y^2\),

\(V=\pi\int_0^2 (y^2)^2\,dy=\pi\int_0^2 y^4\,dy\).

\(V=\pi\left[\dfrac{y^5}{5}\right]_0^2=\dfrac{32\pi}{5}\).

Use \(V=\pi\int x^2\,dy\).

Since \(x=3y\),

\(V=\pi\int_0^2 (3y)^2\,dy=9\pi\int_0^2 y^2\,dy\).

So

\(V=9\pi\left[\dfrac{y^3}{3}\right]_0^2=9\pi\cdot\dfrac{8}{3}=24\pi\).

Use \(V=\pi\int x^2\,dy\).

So

\(V=\pi\int_0^2 (4-y^2)^2\,dy\).

Expand:

\((4-y^2)^2=16-8y^2+y^4\).

Then

\(V=\pi\left[16y-\dfrac{8y^3}{3}+\dfrac{y^5}{5}\right]_0^2\)

\(=\pi\left(32-\dfrac{64}{3}+\dfrac{32}{5}\right)=\dfrac{256\pi}{15}\).

Use \(V=\pi\int y^2\,dx\).

Since \(y=x^{1/3}\), we have \(y^2=x^{2/3}\).

So

\(V=\pi\int_0^1 x^{2/3}\,dx=\pi\left[\dfrac{3}{5}x^{5/3}\right]_0^1=\dfrac{3\pi}{5}\).

Use \(V=\pi\int y^2\,dx\).

First square the expression:

\((x^2-2x+2)^2=x^4-4x^3+8x^2-8x+4\).

So

\(V=\pi\int_0^2 (x^4-4x^3+8x^2-8x+4)\,dx\).

Integrating gives

\(V=\pi\left[\dfrac{x^5}{5}-x^4+\dfrac{8x^3}{3}-4x^2+4x\right]_0^2\)

\(=\pi\left(\dfrac{32}{5}-16+\dfrac{64}{3}-16+8\right)=\dfrac{56\pi}{15}\).