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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 5.2 Volumes

Subsection 5.2.1 Before Class

https://mymedia.ou.edu/media/5.2-1.mp4/1_syajzn32
Figure 46. Pre-Class Video 1
https://mymedia.ou.edu/media/5.2-2/1_xp65khwr
Figure 47. Pre-Class Video 2

Subsubsection 5.2.1.1 Volumes by Cross-Sections

Example 5.2.1.
An isosceles triangle has a base of 10 inches and height of 5 inches. From geometry, we know that its area is 25 in\(^2\text{;}\) use integrals to find its area by slicing vertically, then by slicing horizontally.
Solution.
When slicing vertically, we get the integral \(\ds \int_{-5}^0 x+5\, dx + \int_0^5 -x + 5\, dx\) which gives an area of 25.
When slicing horizontally, we get the integral \(\ds \int_0^5 10-2y\, dy\text{,}\) which also gives an area of 25.
Example 5.2.2.
Describe how to find the volume of the triangular prism below:
A triangular prism with base length 10 inches and height 5 inches; the depth is not specified.
Solution.
Multiply the base triangle area by the length of the prism, so that the volume is \(25L\)
Example 5.2.3.
Show that the volume of a right cylinder of radius \(r\) and height \(h\) has volume \(V = \pi r^2h\)
Solution.
Let \(y\) be the vertical axis. The cross-sectional area is the area of a circle, \(A(y) = \pi r^2\text{.}\) Then, the volume of the solid is
\begin{align*} \ds \int_0^h \pi r^2\, dy \amp = \pi r^2 y\bigg\rvert_0^h \\ \amp = \pi r^2h \end{align*}
For generic volumes, we can consider cross sections of a solid, and “slabs” of constant width.
Definition 5.2.4. Volume.
Let \(S\) be a solid that lies between \(x =a\) and \(x = b\text{.}\) If the cross-sectional area of \(S\) in the plane \(P_x\text{,}\) through \(x\) and perpendicular to the \(x-\)axis is \(A(x)\text{,}\) where \(A\) is a continuous function, then the volume of \(S\) is given by
\begin{equation*} V = \lim_{n\to\infty} \sum_{i=1}^n A(x_i^*)\,\Delta x = \int_a^b A(x)\,dx \end{equation*}
Example 5.2.5.
Show that the volume of a sphere of radius \(r\) is \(V = \dfrac{4}{3}\pi r^3\)
Solution.
Since vertical cross sections are circles of raidus \(r\text{,}\) we have the relationship \(x^2 + y^2 = r^2\text{,}\) so that \(y = \sqrt{r^2-x^2}\text{.}\) This means that the cross-sectional area is given by \(\pi y^2 = \pi (r^2-x^2)\text{.}\) Thus, the volume is given by
\begin{align*} \ds \int_{-r}^r \pi (r^2-x^2)\, dx \amp = \pi r^2 x - \dfrac{1}{3}\pi x^3\bigg\rvert_{-r}^r \\ \amp = \dfrac{4}{3}\pi r^3 \end{align*}

Subsection 5.2.2 Pre-Class Activities

Example 5.2.6.

Draw an explicit connection between how we found area under a curve in Section 4.1 and Section 4.2, and how we’ve defined area here.
Solution.
Answers vary

Example 5.2.7.

Use cross-sectional areas to prove that the volume of a square-based pyramid is \(\dfrac{1}{3}L^2h\text{,}\) where \(L\) is the side length of the square base, and \(h\) is the pyramid’s height.
Solution.
Consider the cross-section with the tip of the pyramid at the origin, and the base at \(x= h\text{.}\) We have a line going through the origin and the point \(\lrpar{h,\dfrac{L}{2}}\text{.}\) This means the equation of the line is given by \(y = \dfrac{L}{2h}x\text{.}\)
Now, cross-sectional areas are squares; this means the height of each square is given by \(2\cdot \dfrac{L}{2h}x = \dfrac{L}{h}x\text{.}\) So, the cross-sectional area is \(\dfrac{L^2}{h^2}x^2\text{.}\) This means the volume is given by \(\ds \int_0^h \dfrac{L^2}{h^2}x^2\, dx = \dfrac{1}{3}L^2h\)

Example 5.2.8.

A wedge is cut out of a circular cylinder of radius 5 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle of \(45^\circ\) along a diameter of the cylinder. Find the volume of the wedge.
Solution.
\(\dfrac{250}{3}\)

Subsection 5.2.3 In Class

Subsubsection 5.2.3.1 Disk Method

Example 5.2.9.
Let \(f(x) = 3\)
  1. Sketch the function on the interval \([0,2]\text{,}\) and find the area between the curve and the \(x-\)axis on \([0,2]\text{.}\)
  2. Sketch the cylinder created by rotating the line segment around the \(x-\)axis; what is the volume of the cylinder?
  3. Write an expression for a slice of the volume of the cylinder.
  4. Create and evaluate an integral expression for the volume of the cylinder.
Solution.
  1. This is the graph of the function \(f(x) = 3\) on the interval \([0,3]\)
  2. This is the sketch of the cylinder created by rotating \(f(x) = 3\) on the interval \([0,2]\) about the \(x-\)axis.
    The volume of the cylinder is \(18\pi\)
  3. \(V = 9\pi \Delta x\text{,}\) where the height of the slice is denoted with \(\Delta x\)
  4. \(\displaystyle \ds \int_0^2 9\pi \, dx = 18\pi\)
Example 5.2.10.
Let \(f(x) = 3x\)
  1. Sketch the region bounded by \(f(x)\text{,}\) \(x=1\text{,}\) and the \(x-\)axis.
  2. Sketch the solid created by rotating the region about the \(x-\)axis on \([0,1]\text{.}\)
  3. Sketch and label an approximating cylinder for the volume of the cone.
  4. Create and evaluate an integral expression to find the volume of the cone.
Solution.
  1. This is the sketch of the \(f(x) = 3x\) on the interval \([0,1.5]\text{.}\)
  2. This is the sketch of the cylinder created by rotating \(f(x) = 3x\) on the interval \([0,1]\) about the \(x-\)axis.
  3. This is the sketch of the cylindrical slice of the solid created by rotating \(f(x) = 3x\) on the interval \([0,1]\) about the \(x-\)axis.
    The radius is \(3x_i\) and the height is \(\Delta x\)
  4. \(\displaystyle \ds \int_0^1 9\pi x^2\, dx = 3\pi\)
Disk Method.
If \(f\) is a positive, continuous function on some interval \([a,b]\text{,}\) then the volume of the solid created by rotating \(f(x)\) about the \(x-\)axis is given by
\begin{equation*} \int_a^b \pi [r(x)]^2\, dx \end{equation*}
where \(r(x)\) is the radius function.
Example 5.2.11.
Find the volume of the solid obtained by rotating the region under the curve \(y = \sqrt{x}\) from 0 to 1, about the \(x-\)axis. Sketch the region and a typical approximating cylinder.
Solution.
This is the sketch of the region under the curve \(y = \sqrt{x}\) on the interval \([0,1]\) that will be rotated about the \(x-\)axis.
This is the sketch of the cylindrical slice of the solid created by rotating \(y = sqrt(x)\) on the interval \([0,1]\) about the \(x-\)axis.
The volume is \(\dfrac{\pi}{2}\)
Example 5.2.12.
Find the volume of the solid obtained by rotating the first-quadrant region bounded by \(y = x^2\text{,}\) \(y = 4\text{,}\) and \(x = 0\) about the \(y-\)axis.
Solution.
\(8\pi\)
Example 5.2.13.
Find the volume of the solid formed by revolving the region between the parabola \(x = y^2 + 1\) and the line \(x = 3\) about the line \(x = 3\text{.}\)
Solution.
\(\dfrac{64\sqrt{2}}{15}\pi\)

Subsubsection 5.2.3.2 Washer Method

Example 5.2.14.
Let \(f(x) = x^2\text{.}\)
  1. Sketch the first-quadrant region bounded by \(f(x)\) and the line \(y = 4\text{.}\)
  2. Sketch the solid resulting from rotating the region about the line \(y = 0\text{.}\)
  3. What are some problems with using the disk method in this example?
Solution.
  1. This is the sketch of the region bounded the curves \(y = x^2\) and \(y = 4\) that will be rotated about the \(x-\)axis.
  2. This is the sketch of the solid formed by rotating the region between the curves \(y = x^2\) and \(y = 4\) about the \(x-\)axis.
  3. The empty space in the middle can’t be captured by a disk.
Example 5.2.15.
Find the volume of the solid formed by rotating the region bounded by \(y = x^2\) and \(y = x\) about the \(x-\)axis.
Solution.
\(\dfrac{128\pi}{5}\)
Washer Method.
If \(f\) and \(g\) are two continuous functions on \([a,b]\) such that \(f(x) \geq g(x)\) on \([a,b]\text{,}\) then the volume of the solid created by rotating the region between \(f\) and \(g\) is given by
\begin{equation*} \int_a^b \pi[(R(x))^2-(r(x))^2]\, dx \end{equation*}
where \(R(x)\) is the outer radius, and \(r(x)\) is the inner radius.
Example 5.2.16.
Find the volume of the solid obtained by rotating the region from the previous example about the line \(y=1\text{.}\)
Solution.
\(\dfrac{\pi}{5}\)
Example 5.2.17.
The region bounded by the curve \(y = x^2 + 1\) and the line \(y = -x + 3\) is revolved about the \(x-\)axis to generate a solid. Find the volume of the solid.
Solution.
\(\dfrac{117\pi}{5}\)
Example 5.2.18.
Find the volume of the solid obtained by rotating the region bounded by the curves \(y = x+1,y=0,x=0,x=2\) about the \(x-\)axis.
Solution.
\(\dfrac{26\pi}{3}\)
Example 5.2.19.
Find the volume of the solid obtained by rotating the region bounded by the curves \(y = \sqrt{25-x^2},y=0,x=2,x=4\) about the \(x-\)axis.
Solution.
\(\dfrac{94\pi}{3}\)

Subsection 5.2.4 After Class Activities

Example 5.2.20.

Find the volume of the solid obtained by rotating the region bounded by the curves \(y = 2\sqrt{x}, y=2, x=0\) about the \(x-\)axis.
Solution.
\(2\pi\)

Example 5.2.21.

Find the volume of the solid obtained by rotating the region bounded by the curves \(y = x^2,x=y^2\) about \(y = 1\text{.}\)
Solution.
\(\dfrac{11\pi}{30}\)

Example 5.2.22.

Find the volume of the solid created by rotating the the region bounded by the curves \(y= x^3\text{,}\) \(y = 0\text{,}\) and \(x = 1\) about \(x = 2\text{.}\)
Solution.
\(\dfrac{3\pi}{5}\)
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