Areas Under Curves onto the y-axis
In calculus, finding the area between a curve and the y-axis is an application integration. This concept extends the idea of finding areas under curves to include regions bounded by the y-axis.
Rather than integrate $y$ with respect to $x$ to determine the area between the $x$-axis and a curve, we can integrate $x$ with respect to $y$ to determine the area with respect to the $y$-axis to a curve.
Setting up the Integral
To find the area between a curve $y = f(x)$ and the y-axis from $y = a$ to $y = b$, we use the integral:
$$ A = \int_a^b x(y) dy $$
Here, $x(y)$ is the function expressing x in terms of y. This is often obtained by solving the original equation $y = f(x)$ for x.
NoteIt's crucial to express the integrand in terms of y, as we're integrating with respect to y.
Example: Area Bounded by a Parabola and the y-axis
ExampleLet's find the area bounded by the parabola $x = y^2$ and the y-axis from $y = 0$ to $y = 2$.
- The function is already in the form $x(y) = y^2$.
- Set up the integral: $A = \int_0^2 y^2 dy$
- Evaluate: $A = [\frac{1}{3}y^3]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$
Therefore, the area is $\frac{8}{3}$ square units.
Common MistakeStudents often forget to change the limits of integration when switching from x to y. Always ensure your limits match your variable of integration!
Volumes of Revolution About the x-axis
When a region bounded by a curve $y = f(x)$, the x-axis, and two vertical lines is rotated around the x-axis, it forms a solid. The volume of this solid can be calculated using integration.
The Disk Method
For a function $y = f(x)$ rotated about the x-axis from $x = a$ to $x = b$, the volume is given by:
$$ V = \pi \int_a^b [f(x)]^2 dx $$
This method imagines the solid as a stack of thin circular disks, each with radius $f(x)$.
TipVisualize each slice as a thin disk with thickness $dx $ and radius $y$ to understand why we square the function inside the integral. We are trying to sum of an infinite number of cylinders with radius $y$ and height $dx$, so the area of each of these cylinders is given by
$$π y^2 dx $$
So the total volume must be :
$$\int π y^2 dx$$
Example: Volume of a Cone
ExampleFind the volume of a cone formed by rotating $y = x$ about the x-axis from $x = 0$ to $x = h$.
- Set up the integral: $V = \pi \int_0^h x^2 dx$
- Evaluate: $V = \pi [\frac{1}{3}x^3]_0^h = \frac{1}{3}\pi h^3$
This gives us the familiar formula for the volume of a cone.
Volumes of Revolution About the y-axis
When rotating a region about the y-axis, we use a similar approach, but we perform the integral
$\pi \int x^2 dy$$
which determines the volume of the shape formed when the curve is rotated about the $y$-axis.
The factor of 2π comes from the circumference of each cylindrical shell, while x represents the radius of each shell.
Example: Volume of a Sphere
ExampleFind the volume of a sphere formed by rotating the semicircle $y = \sqrt{r^2 - x^2}$ about the y-axis from $x = -r$ to $x = r$.
- Set up the integral: $V = 2\pi \int_{-r}^r x\sqrt{r^2 - x^2} dx$
- This integral is complex, but it evaluates to $V = \frac{4}{3}\pi r^3$
This gives us the well-known formula for the volume of a sphere.
Applications in Industrial Design
These techniques find use in industrial design, particularly in:
- Calculating volumes of irregularly shaped containers
- Designing rotational symmetrical objects like vases, bottles, or lampshades
- Optimising material usage in manufacturing processes
A designer is creating a vase whose profile is given by $y = x^2 + 1$ from $x = 0$ to $x = 2$. To calculate the volume of water it can hold, they would use the shell method:
$V = 2\pi \int_0^2 x(x^2 + 1) dx = 2\pi [\frac{1}{4}x^4 + \frac{1}{2}x^2]_0^2 = \frac{22\pi}{3}$ cubic units.
This method allows students to analyse 3-dimensional objects and solve problems relating to the geometry of non-typical shapes, bridging the gap between abstract mathematics and real-world applications.
