Chapter 5: Multiple Integrals

In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between$\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{2}$,

find:

The curved area of this solid.

Above the rectangle with vertices (0,0), (0,1),(2,0),(2,1), and below the surface ${\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{36}{\mathit{x}}^{\mathbf{2}}\mathbf{(}\mathbf{4}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{)}$

Find the gravitational attraction on a unit mass at the origin due to a mass (of constant density) occupying the volume inside the cone

${\mathbf{z}}^{\mathbf{2}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{0}\mathbf{<}\mathbf{z}\mathbf{<}\mathbf{h}$

Above the triangle with vertices (0,0),(2,0), and (2,1), and below the paraboloid $\mathrm{z}=24-{\mathrm{x}}^2-{\mathrm{y}}^2$.

In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between$\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{2}$,

find:

The centroid of the arc.

In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between$\mathbf{x}\mathbf{=}\mathbf{0}$and$\mathit{x}\mathbf{=}\mathbf{2}$,

find:

The centroid of the volume.

Above the triangle with vertices (0,2),(1,1) and (2,2) , and under the surface z = xy.

(a) Find the centroid of the solid paraboloid inside

$\mathbf{z}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{,}\mathbf{0}\mathbf{<}\mathbf{z}\mathbf{<}\mathbf{c}$

(b) Repeat part (a) if the density is$\mathbf{\rho }\mathbf{=}\mathbf{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$

Under the surface z = y(x+2) , and over the area bounded by $\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\sqrt{\mathbf{x}}$.

In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between role="math" localid="1658830478813" $\mathit{x}\mathbf{=}\mathbf{0}$and $\mathit{x}\mathbf{=}\mathbf{2}$,

find:

The centroid of the surface area.