Problem 16 - Calculus and its applications

Question

Find the volume of the solid capped by the surface \(z=x+y\) over the region bounded on the \(x y\) -plane by \(y=1-x, y=0, x=0,\) and \(x=1,\) by evaluating the integral $$ \int_{0}^{1} \int_{0}^{1-x}(x+y) d y d x $$

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{1}{3}\).

Step 1: Define the Region of Integration

The region of integration is bounded by the curves and lines given in the problem: \(y = 1-x\), \(y = 0\), \(x = 0\), and \(x = 1\). This defines a triangular region in the \(xy\)-plane.

Step 2: Set Up the Integral

We are given the double integral \(\int_{0}^{1} \int_{0}^{1-x} (x+y) \ dy \ dx\), which represents the volume under the surface \(z = x + y\) over the defined triangular region.

Step 3: Integrate with Respect to y

The inner integral is \(\int_{0}^{1-x} (x+y) \ dy\). Integrate this with respect to \(y\):\[\int_{0}^{1-x} (x+y) \ dy = \left[xy + \frac{y^2}{2}\right]_{0}^{1-x} = x(1-x) + \frac{(1-x)^2}{2}.\]

Step 4: Evaluate the Inner Integral

Substitute \(y = 1-x\) into the expression from Step 3:\[x(1-x) + \frac{(1-x)^2}{2} = x - x^2 + \frac{1-2x+x^2}{2}.\] Combine similar terms to get:\[x - x^2 + \frac{1}{2} - x + \frac{x^2}{2} = \frac{1}{2} - \frac{x^2}{2}.\]

Step 5: Integrate with Respect to x

Now integrate the result from Step 4 with respect to \(x\):\[\int_{0}^{1} \left(\frac{1}{2} - \frac{x^2}{2}\right) \ dx = \left[ \frac{x}{2} - \frac{x^3}{6} \right]_{0}^{1}.\]

Step 6: Evaluate the Outer Integral

Substitute the limits \(x=0\) and \(x=1\) into the expression from Step 5:\[\left(\frac{1}{2} - \frac{1}{6}\right) - \left(0\right) = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}.\]

Step 7: Conclusion

The volume of the solid capped by the surface is \(\frac{1}{3}\).

Key Concepts

Volume of Solids

When we talk about finding the volume of a solid in calculus, we are typically dealing with shapes that are not as straightforward as cubes or rectangular prisms. Instead, we often have solids that are "capped" by surfaces that can be curvy and irregular. These surfaces can be described with equations, such as the surface in our problem, which is given by the equation \(z = x + y\).
To find the volume of such solids, we use double integrals. This mathematical tool allows us to "add up" infinitely many tiny volumes under the surface to get a total volume. Imagine breaking down the solid into an infinite number of thin slices and adding them all together.
In this exercise, the solution involves integrating the function \(x + y\) over a specific region. This ensures that every part of the solid beneath this surface is accounted for.

Region of Integration

The region of integration refers to the specific portion of the plane over which we are evaluating our integral. It is crucial to clearly define this region as it directly impacts the limits of our integral.
In the given problem, the region is bounded by the lines \(y = 1 - x\), \(y = 0\), \(x = 0\), and \(x = 1\). These boundaries create a triangular shape in the \(xy\)-plane. It is essential to translate this geometric region into correct mathematical limits for the integral.

• We first identify the range for \(x\), which runs from 0 to 1.
• Then we consider the range for \(y\), which varies from 0 to \(1-x\) for a given \(x\). This is because \(y\) is confined by the line \(y = 1-x\) on one side, and by \(y = 0\) on the other.

This setup ensures that our integration covers every point within this triangular region.

Calculus

Calculus provides us with powerful tools like integration to solve problems involving areas and volumes. In this case, we use a double integral to find the volume of the solid.
Integration is all about accumulating values, which in this case, involves accumulating the function \(x + y\) over the triangular region. The double integral \(\int_{0}^{1} \int_{0}^{1-x} (x+y) \,dy \,dx\) sums up these values.

• First, we solve the inner integral \(\int_{0}^{1-x} (x+y) \,dy\), which finds the 'slice' of volume for a given \(x\).
• Next, the result is used in the outer integral \(\int_{0}^{1} \ldots \,dx\), which accounts for all possible \(x\) values within the region.

The final result \(\frac{1}{3}\) represents the total volume, showing how calculus helps us easily manage complex geometric tasks.