Problem 306 - Calculus Volume 3

Question

For the following exercises, express the surface integral as an iterated double integral by using a projection on \(S\) on the yz-plane. \(\iint_{S} x y^{2} z^{3} d S ; S\) is the first-octant portion of plane \(2 x+3 y+4 z=12\)

Step-by-Step Solution

Answer

Convert the surface integral to \[\int_{0}^{4} \int_{0}^{\frac{12-3y}{4}} \frac{12 - 3y - 4z}{2} y^2 z^3 \, dz \, dy.\]

Step 1: Determine the Projection on the yz-plane

The given surface is defined by the plane equation \(2x + 3y + 4z = 12\) in the first octant. To express the surface integral as a double integral using a projection on the yz-plane, first solve for \(x\) in terms of \(y\) and \(z\):\[x = \frac{12 - 3y - 4z}{2}.\] The projection of \(S\) onto the yz-plane is therefore bounded by the points where \(x = 0\), or \(12 - 3y - 4z = 0\). The intercepts and boundaries in the yz-plane occur at \(y = 0, z = 0\), and on the line \(3y + 4z = 12\).

Step 2: Set the Limits for the Double Integral

The region in the \(yz\)-plane is bounded by \(y = 0\), \(z = 0\), and \(3y + 4z = 12\). Thus, the limits for \(y\) and \(z\) are determined by these boundaries. When\(z\) ranges from 0 to \(\frac{12 - 3y}{4}\) and \(y\) from 0 to 4 (since when \(z = 0\), \(y\) can be maximum 4), the validity of the interval is ensured.

Step 3: Set Up the Double Integral

The surface integral \(\iint_{S} x y^2 z^3 \, dS\) can be transformed into a double integral over the projected region. Replace \(x\) using \(\frac{12 - 3y - 4z}{2}\): \[\int_{0}^{4} \int_{0}^{\frac{12-3y}{4}} \frac{12 - 3y - 4z}{2} y^2 z^3 \, dz \, dy.\]

Step 4: Simplify and Evaluate the Double Integral

First, integrate with respect to \(z\):\[\int_{0}^{4} \left(\int_{0}^{\frac{12-3y}{4}} \frac{12 - 3y - 4z}{2} y^2 z^3 \, dz \right) \, dy.\] Break it down further, calculate the inner integral, and keep solving to evaluate the entire integral. This step involves solving the polynomial integral and simplifying the resultant expression over varying bounds.

Key Concepts

Projection on the yz-plane

When you talk about a surface in three-dimensional space, projecting it onto one of the coordinate planes simplifies things. Here, we work with a surface defined by the plane equation \(2x + 3y + 4z = 12\). The goal is to project this surface onto the yz-plane. To do this, solve for \(x\) in terms of \(y\) and \(z\). Start with the equation:

• \[ x = \frac{12 - 3y - 4z}{2} \]

Now, imagine this equation on a two-dimensional plane—the yz-plane. The surface \(S\) is in the first octant, which means all coordinates are non-negative. The projection onto the yz-plane considers the edge where \(x = 0\). When \(x = 0\), the resulting line from \(3y + 4z = 12\) forms the boundary along with the axes \(y = 0\) and \(z = 0\). In this simplified form, we work with a triangular region on the yz-plane defined by these lines, making the math much more manageable.

Iterated Double Integral

Once we project the surface onto the yz-plane, we express the surface integral as a double integral. This iterated double integral evaluates the function over the specific region on the plane. The initial integral expression:

• \( \iint_{S} x y^{2} z^{3} \, dS \)

transforms into a double integral over the yz-plane.By substituting the expression for \(x\) back into the integral, the new iterated form is created:

• \[ \int_{0}^{4} \int_{0}^{\frac{12-3y}{4}} \frac{12 - 3y - 4z}{2} y^2 z^3 \, dz \, dy \]

This process involves separate evaluation—one integral calculated first, followed by the next. The inner integral, done with respect to \(z\), simplifies the function over that variable's bounds. Then, the outer integral completes the evaluation concerning \(y\). It is like peeling an onion; layer by layer, until the entire region is evaluated.

Plane Equation

The fundamental surface under investigation stems from a linear equation involving three variables: \(2x + 3y + 4z = 12\). This equation represents a plane in three dimensions. Understanding plane equations is key in multivariable calculus, as they describe flat, two-dimensional surfaces in three-dimensional space.To work with them effectively:

• Identify the coefficients: They affect the plane's tilt and orientation.
• Recognize the intercepts: The equation intersects the axes when one or two variables are zero. From here, you determine the part lying within the first octant.

For our task, rearranging the equation to solve for one of the variables—\(x\)—helps in projecting onto a plane, simplifying calculations. Each coefficient in the equation changes the slope and intercepts, altering how the plane appears but always follows this balanced form.

Bounds for Double Integral

Setting the correct bounds ensures the integral accounts for the precise region we are evaluating. In this case, these bounds correspond to the edges and intersections of our projected region on the yz-plane.For the exercise, these are:

• \(y = 0\): the lower bound for \(y\).
• \(y = 4\): the upper bound when \(z = 0\).
• \(z = 0\): the lower bound for \(z\).
• \(z = \frac{12 - 3y}{4}\): the form derived from \(3y + 4z = 12\), representing the region’s hypotenuse.

These bounds ensure the integration is confined to the triangular region in the yz-plane. By understanding how these are found, you control the limits over a multi-dimensional space—crucial for precision in math and engineering contexts.