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:

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

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

\(\displaystyle V=\frac{94}{3}\pi\)

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:

\(\displaystyle V=\pi(10^2x-\frac{x^3}{3})|_{-4}^{3}\)

\(\displaystyle V=\pi(10^2(3)-\frac{(3)^3}{3})-\pi(10^2(-4)-\frac{(-4)^3}{3})\)

\(\displaystyle V=\frac{2009}{3}\pi\)

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:

\(\displaystyle V=\pi(3^2x-\frac{x^3}{3})|_{-2}^{1}\)

\(\displaystyle V=\pi(3^2(1)-\frac{(1)^3}{3})-\pi(3^2(-2)-\frac{(-2)^3}{3})\)

\(\displaystyle V=24\pi\)

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:

\(\displaystyle V=\pi(7^2x-\frac{x^3}{3})|_{-6}^{-5}\)

\(\displaystyle V=\pi(7^2(-5)-\frac{(-5)^3}{3})-\pi(7^2(-6)-\frac{(-6)^3}{3})\)

\(\displaystyle V=\frac{56}{3}\pi\)

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:

\(\displaystyle V=\pi(7^2x-\frac{x^3}{3})|_{-3}^{-2}\)

\(\displaystyle V=\pi(7^2(-2)-\frac{(-2)^3}{3})-\pi(7^2(-3)-\frac{(-3)^3}{3})\)

\(\displaystyle V=\frac{128}{3}\pi\)

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:

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

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

\(\displaystyle V=78\pi\)

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:

\(\displaystyle V=\pi(12^2x-\frac{x^3}{3})|_{-6}^{-3}\)

\(\displaystyle V=\pi(12^2(-3)-\frac{(-3)^3}{3})-\pi(12^2(-6)-\frac{(-6)^3}{3})\)

\(\displaystyle V=369\pi\)

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:

\(\displaystyle V=\pi(11^2x-\frac{x^3}{3})|_{-3}^{0}\)

\(\displaystyle V=\pi(11^2(0)-\frac{(0)^3}{3})-\pi(11^2(-3)-\frac{(-3)^3}{3})\)

\(\displaystyle V=354\pi\)

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:

\(\displaystyle V=\pi(2^2x-\frac{x^3}{3})|_{-1}^{1}\)

\(\displaystyle V=\pi(2^2(1)-\frac{(1)^3}{3})-\pi(2^2(-1)-\frac{(-1)^3}{3})\)

\(\displaystyle V=\frac{22}{3}\pi\)

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:

\(\displaystyle V=\pi(6^2x-\frac{x^3}{3})|_{-6}^{-3}\)

\(\displaystyle V=\pi(6^2(-3)-\frac{(-3)^3}{3})-\pi(6^2(-6)-\frac{(-6)^3}{3})\)

\(\displaystyle V=45\pi\)