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:

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

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

\(\displaystyle V=45\pi\)

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:

\(\displaystyle V=\pi(6^2x-\frac{x^3}{3})|_{-3}^{0}\)

\(\displaystyle V=\pi(6^2(0)-\frac{(0)^3}{3})-\pi(6^2(-3)-\frac{(-3)^3}{3})\)

\(\displaystyle V=99\pi\)

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:

\(\displaystyle V=\pi(11^2x-\frac{x^3}{3})|_{2}^{5}\)

\(\displaystyle V=\pi(11^2(5)-\frac{(5)^3}{3})-\pi(11^2(2)-\frac{(2)^3}{3})\)

\(\displaystyle V=324\pi\)

In order to find the volume of this solid, the solid must be formed and observed. 

This solid represents the solid that is formed by taking half a parabola (x² bound by x = 0) and giving it a height of 9 units (bounded by y = 9). This solid forms an object that resembles a vase. The volume of the solid would then be an integral of the area of the cross sections of the vase. Since the vertical cross sections (cutting the vase with vertical planes) give unsymmetrical shapes, it would be better to use horizontal cross sections and change the orientation of the figure to better see how to set up the volume integral.

This orientation can be obtained by changing the axis. This is accomplished by make x the dependent variable, and y the independent variable. To do this, solve the equation for the function in term of x instead of y:

\(\displaystyle y = x^2\)

\(\displaystyle x = \sqrt{y}\)

Now the area of the cross-sections must be calculated. Sample cross sections are shown in the figure above. Since the area of a circle is equal to πr², the differential area expression for this solid would be:

\(\displaystyle A = \pi r^2, r= \sqrt{y}}\)

\(\displaystyle A(y) = \pi y\), now we have the area of the cross section as a function of y

The final step is to find the limits of integration, which in this case, would be from the bottom of the solid to the top. Since it starts at y = 0, and continues to y = 9, those are the limits of integration. To find the volume:

\(\displaystyle V= \int_{a}^{b} A(y)d(y)\)\(\displaystyle V = \int_{0}^{9} \pi y dy= \pi\int_{0}^{9} ydy= \pi \left(\frac{y^2}{2}\right)\)\(\displaystyle V = \pi \left(\frac{9^2}{2}-\frac{0^{2}}{2}\right) = \frac{81\pi}{2}\)