Problem 43
Question
Use the formulas \(m=\int_{C} \rho d s, \bar{x}=\frac{1}{m} \int_{C} x \rho d s\) \(\bar{y}=\frac{1}{m} \int_{c} y \rho d s, I=\int_{C} w^{2} \rho d s.\) Compute the mass \(m\) of a rod with density \(\rho(x, y)=x\) in the shape of \(y=x^{2}, 0 \leq x \leq 3.\)
Step-by-Step Solution
Verified Answer
The mass \(m\) of the rod is \(\frac{9}{2}\).
1Step 1: Setup the integral for mass computation
Since we're working along a rod, the segment lengths, \(ds\), can be replaced with \(dx\). The density function \(\rho(x) = x\). Therefore, the mass can be expressed as the integral of the product of density and the segment length, \(dx\), over the interval from 0 to 3. Mathematically, this is represented as: \[m=\int_{0}^{3} x \, dx.\]
2Step 2: Computation of mass integral
Now, we need to calculate the integral to find the mass. The antiderivative of \(x\) is \(\frac{1}{2}x^{2}\). Hence, applying the Fundamental Theorem of Calculus, we get: \[m= \frac{1}{2}x^{2} \Big|_{0}^{3} = \frac{1}{2}*3^{2} - \frac{1}{2}*0^{2} = \frac{9}{2}.\]
Key Concepts
Integral CalculusDensity FunctionAntiderivative Computation
Integral Calculus
Integral calculus is an essential branch of mathematics that focuses on accumulation and area under curves. In the context of finding the mass of a rod, integral calculus allows us to sum up infinitely small segments to calculate a particular value, like mass. This is done by setting up an integral to integrate the density of the rod over its length.
In this exercise, the rod is defined by a parabola, specifically the curve given by the function \(y = x^2\) over the domain \(0 \leq x \leq 3\). To find the mass \(m\), we set up an integral \(m = \int_{0}^{3} x \, dx\), where \(x\) represents the density function and \(dx\) represents an infinitesimal segment of the rod along the x-axis.
This way, integral calculus helps us compute not only masses but any cumulative measure changing over an interval.
In this exercise, the rod is defined by a parabola, specifically the curve given by the function \(y = x^2\) over the domain \(0 \leq x \leq 3\). To find the mass \(m\), we set up an integral \(m = \int_{0}^{3} x \, dx\), where \(x\) represents the density function and \(dx\) represents an infinitesimal segment of the rod along the x-axis.
This way, integral calculus helps us compute not only masses but any cumulative measure changing over an interval.
Density Function
A density function, often denoted by \( \rho(x) \) (rho of x), describes how mass is distributed along an object, such as a rod in our context. When dealing with variably shaped objects, a uniform density would not suffice to provide accurate representations of mass distributions.
In this exercise, the density is given as \( \rho(x, y) = x \), meaning the density varies linearly with x. This variable density implies that elements further along the x-axis add more mass per unit length than those closer to the origin. Such a function helps in calculating mass by giving us a continuously changing function that adapts over different segments of the rod.
With this, the density function is a key element in integrating to find total values like mass when dealing with variously distributed material properties.
In this exercise, the density is given as \( \rho(x, y) = x \), meaning the density varies linearly with x. This variable density implies that elements further along the x-axis add more mass per unit length than those closer to the origin. Such a function helps in calculating mass by giving us a continuously changing function that adapts over different segments of the rod.
With this, the density function is a key element in integrating to find total values like mass when dealing with variously distributed material properties.
Antiderivative Computation
Antiderivative computation involves finding a function whose derivative matches a given function. This process is crucial in solving integrals, as it allows us to calculate the area under curves or cumulative totals, such as mass.
In the problem at hand, the task is to integrate the function \(x\) to find the mass of the rod. The antiderivative of \(x\) is \(\frac{1}{2}x^2\). Applying the Fundamental Theorem of Calculus, we compute the definite integral for \(x\) between the limits 0 to 3.
This gives us \(m = \frac{1}{2}x^2 \bigg|_{0}^{3} = \frac{1}{2}*3^2 - \frac{1}{2}*0^2 = \frac{9}{2}\). The antiderivative becomes a tool that not only reveals underlying relationships within functions but also applies these relationships to real-world measures like the weight of an object.
In the problem at hand, the task is to integrate the function \(x\) to find the mass of the rod. The antiderivative of \(x\) is \(\frac{1}{2}x^2\). Applying the Fundamental Theorem of Calculus, we compute the definite integral for \(x\) between the limits 0 to 3.
This gives us \(m = \frac{1}{2}x^2 \bigg|_{0}^{3} = \frac{1}{2}*3^2 - \frac{1}{2}*0^2 = \frac{9}{2}\). The antiderivative becomes a tool that not only reveals underlying relationships within functions but also applies these relationships to real-world measures like the weight of an object.
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