Video: Evaluating the Mass of a Non-Uniform Lamina Using Integration

A nonuniform square lamina, 𝐴𝐡𝐢𝐷, has a side length of 15 inches. The density of the lamina across its area from its edge 𝐴𝐡 is given by the function 𝑄(π‘₯) = 1 + sin (2 + 0.02π‘₯Β²) pounds per square inch, where π‘₯ is the distance from the edge 𝐴𝐡 of the lamina. What is the mass, to the nearest pound, of the rectangle remaining after removing the part of the lamina which is greater than 6 inches from the edge 𝐴𝐡?

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Video Transcript

A nonuniform square lamina, 𝐴𝐡𝐢𝐷, has a side length of 15 inches. The density of the lamina across its area from its edge 𝐴𝐡 is given by the function 𝑄 of π‘₯ is equal to one plus the sin of two plus 0.02π‘₯ squared pounds per square inch, where π‘₯ is the distance from the edge 𝐴𝐡 of the lamina. What is the mass, to the nearest pound, of the rectangle remaining after removing the part of the lamina which is greater than six inches from the edge 𝐴𝐡?

This question refers to a nonuniform square lamina 𝐴𝐡𝐢𝐷, which has a side length of 15 inches. A lamina is a two-dimensional planar surface with mass 𝑀 and surface density 𝑄, where 𝑄 may be a constant, a function of π‘₯, or a function of 𝑦, or a function of π‘₯ and 𝑦, in units of mass per area squared, in our case, inches squared. A lamina is said to be uniform if the surface density 𝑄 is constant or nonuniform if 𝑄 is a function of π‘₯, 𝑦, or π‘₯ and 𝑦. For a uniform lamina, the mass 𝑀 is 𝑄, which is the surface density, times the area. And for a nonuniform lamina, the mass 𝑀 is the integral over the area of the surface density function 𝑄.

In our case, we have a nonuniform square lamina, 𝐴𝐡𝐢𝐷, which has side length 15 inches. The density of the lamina across its area from its edge 𝐴𝐡 is given by the function 𝑄 of π‘₯ is equal to one plus the sin of two plus 0.02π‘₯ squared pounds per square inch. π‘₯ in our function is the distance from the edge 𝐴𝐡 of the lamina. Placing our lamina in the π‘₯𝑦-plane, we’d like to find the mass of the rectangle remaining after removing the part of the lamina, which is greater than six inches from the edge 𝐴𝐡. And remember that our mass is the integral of the function 𝑄 over the area. Our function 𝑄 depends on π‘₯ only. And we know our length in the 𝑦-direction is a constant, 15. This means we can multiply our integral by 15 and integrate with respect to π‘₯ only. This gives us the mass is 15 times the integral of our function 𝑄 of π‘₯ with respect to π‘₯.

Now, we need to consider our boundaries for π‘₯ β€” that is the limit of integration. As we’re dealing with a physical object, the density must be positive. So, our square will be in the first quadrant. And the simplest place to put this is with one corner at the origin. Since our rectangle has a side length of six, if our lower limit of Integration is zero, our upper limit is six. We’ll come back and take another look at this once we’ve done our integration and found the mass. So let’s do that now. Our mass is 15 times the integral between zero and six of 𝑄 of π‘₯ with respect to π‘₯, which is 15 times the integral between zero and six of one plus sin two plus 0.02π‘₯ squared with respect to π‘₯. This is not simple to integrate by hand. But we can input this definite integral into a calculator. This gives us a mass of 159.034 pounds, which to the nearest pound is 159.

Let’s just go back and consider our π‘₯ boundaries again. Since our function is a function of π‘₯ only, our 𝑦 is a constant. So, the square can be placed anywhere we like in the vertical direction. To make things simple in the π‘₯-direction, we chose π‘₯ is equal to zero as our lower bound. But we could have put our square anywhere along the π‘₯-axis, say at π‘₯ is equal to π‘Ž. The value we’re putting into our function 𝑄 is the distance from the side 𝐴𝐡 in the π‘₯-direction. So, if the side 𝐴𝐡 is at π‘₯ is equal to π‘Ž, then at π‘₯ is equal to π‘₯ 𝑖, for example, the distance from our side 𝐴𝐡, which is our input, is π‘₯ 𝑖 minus π‘Ž. So, if the side 𝐴𝐡 is at π‘₯ is equal to π‘Ž, we would integrate the function 𝑄 of π‘₯ minus π‘Ž between π‘Ž and π‘Ž plus six.

When we did our calculation, we chose π‘Ž equal to zero. And this gave us a mass of 15 times the integral between zero and six of 𝑄 of π‘₯ with respect to π‘₯. Our mass was, therefore, 159 pounds.

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