Problem 36
Question
Convert the integral $$\int_{-1}^{1} \int_{0}^{\sqrt{1-y^{2}}} \int_{0}^{x}\left(x^{2}+y^{2}\right) d z d x d y$$ to an equivalent integral in cylindrical coordinates and evaluate the result. \(\iiint_{D} f(r, \theta, z) d z r d r d \theta\) over the given region \(D\)
Step-by-Step Solution
Verified Answer
The evaluated integral in cylindrical coordinates is \(\frac{\pi}{5}\).
1Step 1: Understand the Region
The given integral is in Cartesian coordinates where \(x\) ranges from \(0\) to \(\sqrt{1-y^2}\), \(y\) ranges from \(-1\) to \(1\), and \(z\) ranges from \(0\) to \(x\). This region corresponds to a quarter of a circle when rotated around the y-axis.
2Step 2: Convert Limits to Cylindrical Coordinates
In cylindrical coordinates, \(x^2 + y^2 = r^2\). Thus, the circular region from \(-1\) to \(1\) for \(y\) and \(0\) to \(\sqrt{1-y^2}\) for \(x\) is converted to \(r = 0\) to \(1\) and \(\theta = 0\) to \(\pi\). The \(z\) range remains from \(0\) to \(r\).
3Step 3: Convert Function and Evaluate Jacobian
The function \(x^2 + y^2\) becomes \(r^2\) in cylindrical coordinates. The differential \(dx\, dy\, dz\) becomes \(r \, dr \, d\theta \, dz\) due to the cylindrical coordinate conversion.
4Step 4: Set up the New Integral
The integral in cylindrical coordinates becomes: \[\int_{0}^{\pi} \int_{0}^{1} \int_{0}^{r} r^2 \cdot r \, dz \, dr \, d\theta\]Which simplifies to \[\int_{0}^{\pi} \int_{0}^{1} \int_{0}^{r} r^3 \, dz \, dr \, d\theta\]
5Step 5: Evaluate the Integral with respect to \'z\'
First integrate with respect to \(z\): \[\int_{0}^{r} r^3 \, dz = [r^3z]_{0}^{r} = r^4\] Thus, the integral becomes: \[\int_{0}^{\pi} \int_{0}^{1} r^4 \, dr \, d\theta\]
6Step 6: Evaluate the Integral with respect to \'r\'
Now, integrate with respect to \(r\): \[\int_{0}^{1} r^4 \, dr = \left[ \frac{r^5}{5} \right]_{0}^{1} = \frac{1}{5}\]This reduces the integral to: \[\int_{0}^{\pi} \frac{1}{5} \, d\theta\]
7Step 7: Evaluate the Integral with respect to \'\theta\'
Finally, integrate with respect to \(\theta\): \[\int_{0}^{\pi} \frac{1}{5} \, d\theta = \left[ \frac{\theta}{5} \right]_{0}^{\pi} = \frac{\pi}{5}\]
8Step 8: Final Result
The result of the integral in cylindrical coordinates is \(\frac{\pi}{5}\).
Key Concepts
Triple IntegralsCoordinate TransformationVolume IntegrationCalculus Exercises
Triple Integrals
Triple integrals extend the concept of integration to three-dimensional space, allowing us to calculate volumes and other quantities in 3D. In essence, a triple integral operates much like a double integral, but it considers one additional dimension.
Imagine wanting to find the volume of a complex 3D shape. A triple integral allows you to break this shape into infinitesimally small volume elements, sum them up, and determine the total volume. This technique can also be applied to integrals of functions over a volume, such as calculating mass or charge distributions.
Imagine wanting to find the volume of a complex 3D shape. A triple integral allows you to break this shape into infinitesimally small volume elements, sum them up, and determine the total volume. This technique can also be applied to integrals of functions over a volume, such as calculating mass or charge distributions.
- The innermost integral simplifies first, addressing one variable at a time.
- The limits of integration define the region of space to be considered for the computation.
- In our exercise, the triple integral is used to determine the volume of a region bounded in space.
Coordinate Transformation
Switching from Cartesian to cylindrical coordinates is a crucial step in handling certain integrals, particularly those involving circular symmetry.
In cylindrical coordinates, a point in space is represented by a radius, an angle, and a height, denoted as \(r, \theta, z\).
These coordinates simplify the integration process for problems involving cylinders or regions symmetric about an axis. Conversion formulas for such transformations are:
This multiplier accounts for the "stretching" of space as you move to polar dimensions. It is essential for converting area elements correctly when performing integration.
In cylindrical coordinates, a point in space is represented by a radius, an angle, and a height, denoted as \(r, \theta, z\).
These coordinates simplify the integration process for problems involving cylinders or regions symmetric about an axis. Conversion formulas for such transformations are:
- \(x = r \cos(\theta)\)
- \(y = r \sin(\theta)\)
- \(z = z\)
This multiplier accounts for the "stretching" of space as you move to polar dimensions. It is essential for converting area elements correctly when performing integration.
Volume Integration
Volume integration is a powerful technique to compute the size or volume of a solid in space. This is particularly relevant in physics and engineering, where understanding the distribution of mass, density, or other variables within a volume is key.Converting the integral into cylindrical coordinates was specifically beneficial because the original region described a cylinder segment.
With cylindrical integration:
With cylindrical integration:
- The radial limits \( r = 0 \) to \( r = 1 \) concern the distance from the origin to the edge of the circle.
- The angular limits \( \theta = 0 \) to \( \pi \) cover half the circle, ensuring only the specified region is calculated.
- The height limits \( z = 0 \) to \( z = r \) reflect the constraints given by the initial Cartesian parameters.
Calculus Exercises
Through calculus exercises like the one discussed, students learn essential skills for solving real-world problems. They present opportunities to apply theoretical concepts like triple integrals and coordinate transformation practically.
Engaging with such exercises helps to:
- Enhance spatial reasoning by translating geometric regions into mathematical models.
- Develop a comprehensive understanding of integrating functions over volumes.
- Foster the ability to manipulate mathematical expressions through variable change.
Other exercises in this chapter
Problem 36
Evaluate \(\int_{-1}^{1} \int_{0}^{\pi / 2} x \sin \sqrt{y} d y d x\)
View solution Problem 36
Sketch the region of integration and write an equivalent double integral with the order of integration reversed. \begin{equation} \int_{0}^{1} \int_{1-x}^{1-x^{
View solution Problem 36
Find the volumes of the regions in Exercises \(23-36\) $$ \begin{array}{l}{\text { The region bounded in back by the plane } x=0, \text { on the front and }} \\
View solution Problem 37
Use Fubini's Theorem to evaluate $$\int_{0}^{2} \int_{0}^{1} \frac{x}{1+x y} d x d y$$
View solution