Mass of a plane figure bounded by y = 8 − x² and y = x + 2 with density ρ = 2x²

This problem asks to find the mass of a plane figure with variable density using a double integral.

Problem

The figure is bounded by the curves:

y = 8 - x^{2}, y = x + 2, ρ (x, y) = 2 x^{2}

Solution

Step 1

Find the intersection points of the curves. To do this, equate the expressions for y:

Why do we need the intersection points?

The intersection points determine the bounds of the integration region along x. Without them, it is impossible to set the limits in the outer integral correctly.

8 - x^{2} = x + 2 \Rightarrow x^{2} + x - 6 = 0 \Rightarrow x_{1} = - 3, x_{2} = 2

Step 2

Find the corresponding values of y by substituting x into the equation of the line:

A (- 3; - 1), B (2; 4)

Step 3

Determine the integration region D. The parabola y = 8 − x² lies above the line y = x + 2 on the segment [−3; 2] (check: at x = 0 the parabola gives 8, the line gives 2). Hence the region D is given by the inequalities:

D : - 3 \leq x \leq 2, x + 2 \leq y \leq 8 - x^{2}

Why is the parabola above the line?

The vertex of the parabola y = 8 − x² is at (0; 8), with branches pointing downward. The line y = x + 2 passes through (0; 2) with slope 1. On the segment between the intersection points, the parabola lies above the line, which is easily verified by substituting x = 0.

Step 4

The mass of a plane figure with variable density ρ(x, y) is given by:

m = \iint_{D} ρ (x, y) d x d y = \int_{- 3}^{2} d x \int_{x + 2}^{8 - x^{2}} 2 x^{2} d y

Step 5

Evaluate the inner integral over y (for fixed x the density 2x² is moved outside the integral sign):

\int_{x + 2}^{8 - x^{2}} 2 x^{2} d y = 2 x^{2} (6 - x - x^{2})

Step 6

After substituting the limits, we obtain the outer integral over x:

m = \int_{- 3}^{2} (12 x^{2} - 2 x^{3} - 2 x^{4}) d x

Step 7

Evaluate the definite integral using the Newton–Leibniz formula:

m = {[4 x^{3} - \frac{x^{4}}{2} - \frac{2 x^{5}}{5}]}_{- 3}^{2} = \frac{56}{5} - (- \frac{513}{10}) = \frac{125}{2}

Answer

m = 125/2.

Frequently asked questions

How do I find the mass of a plane figure using a double integral?

The mass is computed by m = ∬_D ρ(x, y) dx dy, where D is the region occupied by the figure and ρ(x, y) is the surface density. To evaluate it, the double integral is reduced to an iterated one by setting the limits along the boundaries of the region.

How do I determine the order of traversing the region?

First, find the intersection points of the boundary curves — they set the limits along x. Then, for each x, determine which curve lies below and which lies above — these are the limits along y. In our case −3 ≤ x ≤ 2 and x + 2 ≤ y ≤ 8 − x².

Why is the density moved outside the inner integral?

The density ρ(x, y) = 2x² depends only on x. When integrating over y (with x fixed), the factor 2x² is a constant and hence can be moved outside the integral sign.

What if the boundary curves intersect at more than two points?

If the curves intersect at more than two points, the region is split into several parts and the integral is computed on each separately, then the results are added. In our case the parabola and the line intersect at exactly two points, so the region is single.