Question

(a) Sketch the region \(R\) given by the problem. (b) Set up the iterated integrals, in both orders, that evaluate the given double integral for the described region \(R\) (c) Evaluate one of the iterated integrals to find the signed volume under the surface \(z=f(x, y)\) over the region \(R .\) \(\iint_{R} y e^{x} d A,\) where \(R\) is bounded by \(x=0, x=y^{2}\) and \(y=1 .\)

Short answer

The signed volume is \( \frac{e - 2}{2} \).

Step-by-step solution

Understand the Region R

Region \( R \) is defined by two curves and a line: the parabola \( x = y^2 \), the line \( x = 0 \), and the horizontal line \( y = 1 \). We need to understand the area enclosed by these curves on the \(xy\)-plane.

Sketch Region R

First, plot the parabola \( x = y^2 \), which opens to the right. Then, mark the line \( x = 0 \), which is the y-axis. Finally, draw the horizontal line \( y = 1 \). The region \( R \) is bounded to the right by the parabola, to the left by the y-axis, and from above by \( y = 1 \).

Set Up Iterated Integral (Order: dx dy)

For the order \( dx \, dy \), \( x \) varies from \( 0 \) to \( y^2 \) (left to right), and \( y \) varies from \( 0 \) to \( 1 \). This gives the integral:\[\int_{0}^{1} \int_{0}^{y^2} y e^x \, dx \, dy\]

Set Up Iterated Integral (Order: dy dx)

For the order \( dy \, dx \), \( y \) varies from \( 0 \) to \( \sqrt{x} \) (top to bottom), and \( x \) ranges from \( 0 \) to \( 1 \) (limits of the parabola). This gives the integral:\[\int_{0}^{1} \int_{0}^{\sqrt{x}} y e^x \, dy \, dx\]

Evaluate Iterated Integral (Order: dx dy)

Evaluate the inner integral first:\[\int_{0}^{y^2} y e^x \, dx = y \left[e^x\right]_{0}^{y^2} = y(e^{y^2} - 1)\]Substitute this result into the outer integral:\[\int_{0}^{1} y (e^{y^2} - 1) \, dy\]This integral can be split into two parts:\[\int_{0}^{1} y e^{y^2} \, dy - \int_{0}^{1} y \, dy\]Evaluate these integrals:For the first integral, use substitution, let \( u = y^2\), then \( du = 2y \, dy\) gives:\[\frac{1}{2} \int_{0}^{1} e^u \, du = \frac{1}{2}(e^1 - e^0) = \frac{1}{2}(e - 1)\]For the second integral:\[\int_{0}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1}{2}\]Combine results:\[\frac{e - 1}{2} - \frac{1}{2} = \frac{e - 2}{2}\]

Final Result

The signed volume under the surface \( z = f(x, y) = y e^x \) over region \( R \) is \( \frac{e - 2}{2} \).

Key concepts

Iterated Integrals

When dealing with double integration, iterated integrals are a foundational concept. They allow us to evaluate the integral over a two-dimensional region by breaking it down into two separate integrals. For the problem at hand, we need to set up two iterated integrals: one as \(dx \, dy\) and another as \(dy \, dx\).

• The order \(dx \, dy\) means that we integrate with respect to \(x\) first while keeping \(y\) constant. The limits for \(x\) typically depend on \(y\).
• The reverse order \(dy \, dx\) involves integrating with respect to \(y\) first, then using the results in the \(x\) integration phase.

This approach simplifies complex regions by integrating over a single variable, reducing the overall work before combining the results.

Region Sketching

To successfully set up iterated integrals, accurate region sketching is imperative. For region \(R\), it involves identifying the area bounded by the curves: \(x = y^2\), \(x = 0\), and \(y = 1\).

• The parabola \(x = y^2\) opens to the right and reflects the symmetry along the y-axis.
• The line \(x = 0\) serves as the y-axis, being a constant vertical limit.
• The horizontal line \(y = 1\) caps the region, initiating y-bound integration.

By accurately sketching, you define the incremental areas for integration to occur, creating bounds for either direction of iteration.

Signed Volume

In the realm of multivariable calculus, signed volume represents the net volume above and possibly below the xy-plane. When we evaluate an iterated integral, it often results in a signed area or a signed volume. For this exercise, computing the surface integral \( \iint_{R} y e^{x} \, dA\) provides the total signed volume under the surface \(z = y e^x\) over the defined region \(R\). The result \( \frac{e-2}{2} \) is obtained through:

• Integrating the function over the sketched region \(R\), giving components in both positive and negative directions.
• Revealing how surface overlaps and voids in regions can affect the volumetric total.

Substitution Method

One of the powerful methods to solve the given iterated integral is the substitution method. This technique involves changing the variable to simplify the integration process. In our case, we utilize substitution when evaluating the integral:

• Let \(u = y^2\), then \(du = 2y \, dy\), allowing the substitution instead of directly solving \( y e^{u/2}\).
• Transforming the integral enables direct application of known integral forms, such as exponential integration.

The substitution streamlines solving intricate integrals by converting them to more tractable forms, especially helpful in multivariable calculus contexts.