Question

A triangular lamina is bounded by the coordinate axes and the line \(x+y=6\). Find its mass if its density at each point \(P\) is proportional to the square of the distance from the origin to \(P\).

Short answer

216k

Step-by-step solution

Determine the equation of the density function

The line equation is given by: \( x + y = 6 \). The density at each point is proportional to the square of the distance from the origin. The distance from the origin to a point \((x, y)\) is given by \( \sqrt{x^2 + y^2} \), so the density function \( \rho(x, y) \) will be: \( \rho(x, y) = k(x^2 + y^2) \), where \(k\) is the proportionality constant.

Set up the double integral for mass

The mass of the lamina will be calculated using the double integral of the density function over the region bounded by the line and the coordinate axes. The region can be described by the limits: 0 \leq x \leq 6 and 0 \leq y \leq 6 - x. The mass \(M\) is given by: \( M = \int_0^6 \int_0^{6-x} k(x^2 + y^2) \, dy \, dx \).

Integrate with respect to y

Evaluate \(\int_0^{6-x} (x^2 + y^2) \, dy\): \( \int_0^{6-x} x^2 \, dy + \int_0^{6-x} y^2 \, dy\).First integral: \( x^2 \int_0^{6-x} \, dy = x^2 [ y ]_0^{6-x} = x^2(6-x) \). Second integral: \( \int_0^{6-x} y^2 \, dy = [ \frac{y^3}{3} ]_0^{6-x} = \frac{(6-x)^3}{3} \). Therefore, the result of the integration is: \((6-x)x^2 + \frac{(6-x)^3}{3} \).

Integrate with respect to x

Evaluate \( k \int_0^6 [(6-x)x^2 + \frac{(6-x)^3}{3}] \, dx \). This can be separated into two integrals: \( k \int_0^6 x^2(6-x) \, dx + k \int_0^6 \frac{(6-x)^3}{3} \, dx \).

Solve the first integral

Evaluate \( k \int_0^6 x^2(6-x) \, dx \). Expand and integrate: \( k \int_0^6 (6x^2 - x^3) \, dx = k [ 6 \frac{x^3}{3} - \frac{x^4}{4} ]_0^6 \). Calculate: \( k [ 2x^3 - \frac{x^4}{4}]_0^6 = k[ 2(6)^3 - \frac{(6)^4}{4}] - (0) = k [ 432 - 324 ] = k \cdot 108 \).

Solve the second integral

Evaluate \( k \frac{1}{3} \int_0^6 (6-x)^3 \, dx \). Let \( u = 6 - x \), then \(du=-dx\).Integrate: \( -k \frac{1}{3} \int_6^0 u^3 \, (-du) \). Calculate: \( k \frac{1}{3} \int_0^6 u^3 \, du = k \frac{1}{3} [ \frac{u^4}{4} ]_0^6 = k \frac{1}{3} \frac{(6)^4}{4} = k \frac{1296}{12} = k \cdot 108 \).

Combine results and find the mass

Add the results from Step 5 and Step 6: \( k \cdot 108 + k \cdot 108 = 216k \). Thus, the mass of the triangular lamina is: \( M = 216k \).

Key concepts

density function

In this problem, determining the mass of a triangular lamina relies on understanding its density function. The density of a lamina measures how much mass exists per unit area at any point in the region. For this triangle, the density at each point is proportional to the square of the distance from the origin.To express this mathematically, consider a point \((x, y)\) on the lamina. The distance from the origin \((0, 0)\) to this point is given by \(\root{x^2 + y^2}\). Since the density is proportional to the square of the distance, our density function \(\rho(x, y) = k(x^2 + y^2)\), where \(k\) is a constant of proportionality. This function helps us set up the integral needed for finding the mass. It's important because it tailors the mass calculation to vary depending on how far into the triangle you are.

double integral

The mass of the triangular lamina is calculated using a double integral. A double integral allows us to add up, or integrate, values over a two-dimensional region. In this case, we use it to integrate the density function over the triangular area.The mass \(M\) of the lamina is represented as: \[M = \root{0}{6} \root{0}{6-x} k(x^2 + y^2) dy dx\] The integral first integrates the variable \(y\) within inner limit bounds \((0 \to 6-x)\) and then integrates the variable \(x\) from \((0 \to 6)\). The density function \(k(x^2 + y^2)\) is multiplied at each point. This step-by-step integration helps accumulate the mass correctly throughout the whole triangular region.

proportional density

The concept of proportional density is key to this problem. The density being proportional to the square of the distance from the origin means that the density increases as you move away from the origin.Mathematically, we express this with \( \rho(x, y) = k(x^2 + y^2)\). Here's how to break it down:

• Proportionality Constant: \(k\) adjusts how density scales with distance. It's a factor to ensure the units and magnitudes are correct.
• Square of Distance: \(x^2 + y^2\) shows that as either \(x\) or \(y\) increases, the distance and thus the density increase quadratically.

The proportional density makes the mass non-uniform, meaning parts of the lamina farther from the origin hold more mass.

triangular region integration

To find the mass of the lamina, we perform integration over the triangular region. The region is bound by the coordinate axes (x >= 0, y >= 0) and the line \((x + y = 6)\).When setting up integrals over such regions, it helps to visualize the boundaries. Here’s how it's structured:

• x varies from 0 to 6 (from the y-axis to where \(x = 6\)).
• For any value of \(x\), y varies from 0 up to \(6 - x\) (on the line from intercept to intercept).

This structured boundary ensures the integration covers every point within the triangle. Such integration divides the triangular area into infinitesimally small parts, sums the density at those parts, achieving accurate mass calculation.

distance from origin

The distance from the origin to any point \( (x, y) \) in a plane is given by the formula \(\root{x^2 + y^2}\). It's an application of the Pythagorean theorem. This distance is critical in our problem since the density function is reliant on it. Here’s why it matters:The square of this distance, denoted as \(( x^2 + y^2 )\), measures how far each point on the lamina is from the origin.

• At \( (0,0) \), the distance is 0, meaning density is lowest at the origin.
• As you move away, the distance increases and so does the density, making parts of the triangle farthest from the origin denser.

Considering this distance ensures that we accurately model and integrate the mass variation throughout the triangular region, leading to our final mass calculation, \(M = 216k\).