To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius $\displaystyle r$ in the Cartesian coordinate system is given as:

$\displaystyle x^2+y^2=r^2$

Which can be rewritten in terms of $\displaystyle y$ as $\displaystyle y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$\displaystyle V=\int_{a}^{b}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\displaystyle \pi hr_{cylinder}^2$ where $\displaystyle r_{cylinder}^2\rightarrow f^2(x)$ and $\displaystyle h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating $\displaystyle y$ as our $\displaystyle f(x)$, this integral can be written as:

$\displaystyle V=\int_{a}^{b}\pi(r^2-x^2)dx$

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of $\displaystyle x$ on the circle:

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$\displaystyle V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$\displaystyle V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$\displaystyle V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$\displaystyle V=\pi(13.5^2x-\frac{x^3}{3})|_{-1.8}^{3.2}$

$\displaystyle V=\pi(13.5^2((3.2)-(-1.8))-\frac{(3.2)^3-(-1.8)^3}{3})=2822.4$

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius $\displaystyle r$ in the Cartesian coordinate system is given as:

$\displaystyle x^2+y^2=r^2$

Which can be rewritten in terms of $\displaystyle y$ as $\displaystyle y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$\displaystyle V=\int_{a}^{b}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\displaystyle \pi hr_{cylinder}^2$ where $\displaystyle r_{cylinder}^2\rightarrow f^2(x)$ and $\displaystyle h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating $\displaystyle y$ as our $\displaystyle f(x)$, this integral can be written as:

$\displaystyle V=\int_{a}^{b}\pi(r^2-x^2)dx$

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of $\displaystyle x$ on the circle:

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$\displaystyle V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$\displaystyle V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$\displaystyle V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$\displaystyle V=\pi(25^2x-\frac{x^3}{3})|_{-23}^{-3}$

$\displaystyle V=\pi(25^2((-3)-(-23))-\frac{(-3)^3-(-23)^3}{3})=\frac{25360}{3}\pi$

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius $\displaystyle r$ in the Cartesian coordinate system is given as:

$\displaystyle x^2+y^2=r^2$

Which can be rewritten in terms of $\displaystyle y$ as $\displaystyle y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$\displaystyle V=\int_{a}^{b}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\displaystyle \pi hr_{cylinder}^2$ where $\displaystyle r_{cylinder}^2\rightarrow f^2(x)$ and $\displaystyle h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating $\displaystyle y$ as our $\displaystyle f(x)$, this integral can be written as:

$\displaystyle V=\int_{a}^{b}\pi(r^2-x^2)dx$

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of $\displaystyle x$ on the circle:

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$\displaystyle V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$\displaystyle V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$\displaystyle V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$\displaystyle V=\pi(5^2x-\frac{x^3}{3})|_{-3}^{4}$

$\displaystyle V=\pi(5^2((4)-(-3))-\frac{(4)^3-(-3)^3}{3})=\frac{434}{3}\pi$

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius $\displaystyle r$ in the Cartesian coordinate system is given as:

$\displaystyle x^2+y^2=r^2$

Which can be rewritten in terms of $\displaystyle y$ as $\displaystyle y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$\displaystyle V=\int_{a}^{b}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\displaystyle \pi hr_{cylinder}^2$ where $\displaystyle r_{cylinder}^2\rightarrow f^2(x)$ and $\displaystyle h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating $\displaystyle y$ as our $\displaystyle f(x)$, this integral can be written as:

$\displaystyle V=\int_{a}^{b}\pi(r^2-x^2)dx$

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of $\displaystyle x$ on the circle:

$\displaystyle V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$\displaystyle V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$\displaystyle V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$\displaystyle V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$\displaystyle V=\pi(7.2^2x-\frac{x^3}{3})|_{-1.8}^{3.6}$

$\displaystyle V=\pi(7.2^2((3.6)-(-1.8))-\frac{(3.6)^3-(-1.8)^3}{3})=824.5$

The surface area of a cylinder formed by this region is $\displaystyle A(x)=2\pi(radius)(height)$, where $\displaystyle radius=x$ and $\displaystyle height = (x-1)(x-3)^{2}$. This gives us $\displaystyle A(x)= 2\pi(x)((x-1)(x-3)^{2})=2\pi(x^{4}-7x^{3}+15x^{2}-9x)$.

To obtain the volume, we integrate over this area, finding the bounds where $\displaystyle (x-1)(x-3)^{2}=0$, at x=1 and x=3. Integrating, we get:

$\displaystyle V=\int_1^32\pi(x^{4}-7x^{3}+15x^{2}-9x)dx=2\pi\left [ \frac{x^{5}}{5}-\frac{7x^{4}}{4}+5x^{3}-\frac{9x^{2}}{2}\right ]_1^3=\frac{24\pi}{5}$

We can relabel the curve $\displaystyle y=\sqrt[3]{x}$ as $\displaystyle x=y^{3}$.

The radius of a single shell can be expressed as $\displaystyle y$ and the width can be expressed as $\displaystyle 8-y^{3}$, giving a cross sectional area of $\displaystyle A(y)=2\pi(8y-y^{4})$.

To obtain the volume of solid, we integrate over each section. The first cylinder will cut into the solid at $\displaystyle y=0$ and the last cylinder will intersect at $\displaystyle y=2$.

Integrating, we get: 

$\displaystyle V=2\pi\int_0^2(8y-y^{4})dy= 2\pi\left [ (4y^2-\frac{y^{5}}{5}) \right ]_0^2=\frac{96\pi}{5}$

$\displaystyle V=2\pi\int_0^3((x+1)-(9-x^{2}))dx=2\pi\int_0^3(-x^{3}-x^{2}+9x+9)dx=2\pi \left [ (-\frac{x^{4}}{4}-\frac{x^{3}}{3}+\frac{9x^{2}}{2}+9x) \right ]_0^3=\frac{153\pi}{2}$

Using $\displaystyle x=y^{2}$ and $\displaystyle x=2-y$, we find that the graphs intersect at (1,1).

The sample cylindrical shell's height is $\displaystyle h=2-y-y^{2}$ and the radius is $\displaystyle r=y$.

Integrating, we get:

$\displaystyle V=2\pi\int_0^1(y(2-y-y^{2}))dy=2\pi\left [ y^{2}-\frac{y^{3}}{3}-\frac{y^{4}}{4}\right ]_0^1=\frac{5\pi}{6}$

The parabola's x-intercepts are 0 and 1. The radius is $\displaystyle 2-x$, the circumference is $\displaystyle 2\pi(2-x)$ and the height is $\displaystyle (x-x^{2})$.

Integrating, we get:

$\displaystyle V=\int_0^1(2\pi)(2-x)(x-x^{2})dx=2\pi\int_0^1(x^{3}-3x^{2}+2x)dx \newline =2\pi\left [ \frac{x^{4}}{4}-x^{3}+x^{2}\right ]_0^1 = \frac{\pi}{2}$

The radius of a single shell can be expressed as $\displaystyle x$, the circumference can be expressed as $\displaystyle 2\pi x$, and the height can be expressed as $\displaystyle x-x^{2}$. 

To obtain the volume of all the shells, we integrate the overall volume.  Our bounds can be found by calculating the endpoints, or where $\displaystyle x=x^{2}$. Thus, our endpoints are at $\displaystyle x=0$ and $\displaystyle x=1$.

Integrating, we get: 

$\displaystyle V= \int_0^1 (2\pi x)(x-x^{2})dx=2\pi\int_0^1(x^{2}-x^{3})dx =2\pi\left [ \frac{x^{3}}{3}-\frac{x^4}{4}\right ]_0^1=\frac{\pi}{6}$