Disc method worksheet

If we rotate the enclosed region about the $x$‍-axis, the resulting solid will have no holes or gaps. So we can find the volume of the solid using the disk method.

The image below shows a disk cross section of the solid.

The radius of each disk is $r=\sqrt{x}$‍, and so the area of the face of each disk is $A(x)=\pi r^2=\pi (\sqrt{x}{)}^2=\pi x$‍.

The volume of each disk can be found by multiplying the area of the face by the width of the disk, $dx$‍, and so $v(x)=\pi xdx$‍.

We want to sum up the volumes of infinitely many of these disks between $x=1$‍ and $x=4$‍, and so the volume of the entire solid is given by $V(x)={\displaystyle {\int }_1^4\pi xdx.}$‍

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{15\pi }{2}}$‍cubic units.

This problem is similar to problem 1, except we aren't specifically given the left and right endpoints of the enclosed region. To find these, we need to find where the curve $y=x^2-2x$‍ intersects with the $x$‍-axis.

A sketch of the given region and one representative disk is shown below.

We see that the radius of each disk is $r=-(x^2-2x)$‍. (Note that the negative is needed here to make the radius a positive value.) So it follows that the area of the face of each disk is $A(x)=\pi (-(x^2-2x){)}^2$‍.

To find the volume, we multiply the area of the face by the width of the disk, $dx$‍, and so the volume of each disk is $v(x)=\pi (-(x^2-2x){)}^{2}dx$‍.

We want to sum up the volumes of infinitely many of these disks between $x=0$‍ and $x=2$‍, and so the volume of the entire solid is given by $V(x)={\displaystyle {\int }_0^2(-(x^2-2x){)}^{2}dx}$‍

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{16\pi }{15}}$‍cubic units.

In this problem, the enclosed region is not rotated about the $x$‍-axis, but instead about a line parallel to the $x$‍-axis, $y=4$‍.

Let's sketch this region and a cross section of the solid. Notice that $y={\displaystyle \frac{1}{2}}{x}^3$‍ intersects $y=4$‍, when $x=2$‍.

Now we can see that the radius of each disk is $r=4-{\displaystyle \frac{1}{2}}{x}^3$‍. So the area of the face of each disk is $A(x)=\pi {(4-{\displaystyle \frac{1}{2}}{x}^3)}^2$‍, and the volume of each disk is $v(x)=\pi {(4-{\displaystyle \frac{1}{2}}{x}^3)}^{2}dx$‍, where $dx$‍ is the width of the disk.

The definite integral $V(x)={\displaystyle {\int }_0^2\pi {(4-{\displaystyle \frac{1}{2}}{x}^3)}^{2}dx}$‍ gives the volume of the entire solid.

We evaluate the integral below.

The volume of the solid is ${\displaystyle \frac{144\pi }{7}}$‍cubic units.

In this problem, note that the enclosed region is rotated about the $y$‍-axis not the $x$‍-axis.

Let's sketch this region and a cross section of the solid.

Here the disks are horizontal, and so the width of each disk is $dy$‍. Therefore, we should find all of our components in terms of $y$‍.

From the image, we see that the radius of each disk is $x$‍, but since $x=\sqrt{16-y^2}$‍, it follows that $r=\sqrt{16-y^2}$‍.

We can now express the area of the face of each disk as $A(y)=\pi (\sqrt{16-y^2}{)}^2$‍, and the volume of each disk as $v(y)=\pi (\sqrt{16-y^2}{)}^{2}dy$‍.

To find the volume of the entire solid, we sum up the volumes of infinitely many disks between $y=0$‍ and $y=4$‍. The definite integral $V(y)={\displaystyle {\int }_0^4\pi (\sqrt{16-y^2}{)}^{2}dy}$‍ gives the volume of the entire solid.

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{128\pi }{3}}$‍cubic units.

We start by sketching the enclosed region and a representative cross section of our solid.

Here, the disks are horizontal, and so the width of each disk is $dy$‍. Therefore, we should find all of our components in terms of $y$‍.

The radius of each disk is $r=x$‍. However, expressed in terms of $y$‍, we see that the radius is $r=y^{\frac{3}{2}}$‍. The process is shown below:

So it follows that the area of a face of one disk is $A(y)=\pi {\left(y^{\frac{3}{2}}\right)}^2=\pi y^3$‍ and the volume of one disk is $v(y)=\pi y^{3}dy$‍.

We want to sum up the volumes of infinitely many of these disks between $y=0$‍ and $y=1$‍, and so the volume of the entire solid is given by $V(y)={\displaystyle {\int }_0^1\pi y^{3}dy}$‍

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{\pi }{4}}$‍cubic units.

In this problem, the enclosed region is not rotated about the $y$‍-axis, but instead about a line parallel to the $y$‍-axis, $x=9$‍.

Let's sketch this region and a cross section of the solid.

Notice that the disks are horizontal, and so the width of each disk is $dy$‍. Therefore, we should find all of our components in terms of $y$‍.

Since $y=\sqrt{x}$‍, it follows that $x=y^2$‍.

From the image, we see that the radius of a disk is $9-x$‍. Expressing this in terms of $y$‍, we have that $r=9-y^2$‍.

So the area of the face of each disk is $A(y)=\pi {(9-y^2)}^2$‍, and the volume of each disk is $v(y)=\pi {(9-y^2)}^{2}dy$‍.

We want to sum up the volumes of infinitely many of these disks between $y=0$‍ and $y=3$‍, and so the volume of the entire solid is given by $V(y)={\displaystyle {\int }_0^3\pi {(9-y^2)}^{2}dy}$‍

We can evaluate the definite integral to find the volume of the solid.

The volume of the solid is ${\displaystyle \frac{648\pi }{5}}$‍cubic units.