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 \(R .\) \(\iint_{R} x^{2} y d A,\) where \(R\) is bounded by \(y=\sqrt{x}\) and \(y=x^{2} .\)

Short answer

The signed volume is \( \frac{3}{56} \).

Step-by-step solution

Understand the Region R in the xy-plane

The region \( R \) is bounded by the curves \( y = \sqrt{x} \) and \( y = x^2 \). First, understand that \( y = \sqrt{x} \) is a curve where we take the square root of \( x \) for values \( x \geq 0 \), and \( y = x^2 \) is a parabola opening upwards. To find their intersection points, solve \( \sqrt{x} = x^2 \).

Find Intersection Points

Equate \( y = \sqrt{x} \) and \( y = x^2 \) to find their intersection. \( \sqrt{x} = x^2 \) becomes \( x = x^4 \). Solve for \( x \): \( x(x^3 - 1) = 0 \) gives \( x = 0 \) and \( x = 1 \) (since \( x^3 = 1 \) implies \( x = 1 \)). The curves intersect at points \( (0,0) \) and \( (1,1) \).

Visualize and Sketch the Region

Visualize the region between the curves within the intersection limits. From \( x = 0 \) to \( x = 1 \), the curve \( y = \sqrt{x} \) is above \( y = x^2 \). Sketch these curves to show region \( R \), "sandwiched" between \( y = \sqrt{x} \) (top) and \( y = x^2 \) (bottom).

Write Iterated Integrals in Both Orders

First, write the integral with \( dy \) first, then \( dx \):\[ \int_{0}^{1} \int_{x^2}^{\sqrt{x}} x^2y \ dy \ dx \]Now, write with \( dx \) first, then \( dy \): For each \( y \), \( x \) ranges from \( \sqrt{y} \) to \( y^{1/2} \):\[ \int_{0}^{1} \int_{y^2}^{\sqrt{y}} x^2y \ dx \ dy \]

Choose an Integral and Evaluate

Let’s evaluate \( \int_{0}^{1} \int_{x^2}^{\sqrt{x}} x^2y \ dy \ dx \).Start by integrating \( x^2y \) with respect to \( y \):\[ \int_{x^2}^{\sqrt{x}} x^2y \ dy = x^2 \left[ \frac{y^2}{2} \right]_{x^2}^{\sqrt{x}} = x^2 \left( \frac{(\sqrt{x})^2}{2} - \frac{(x^2)^2}{2} \right) = x^2 \left( \frac{x}{2} - \frac{x^4}{2} \right) \]\[ = \frac{x^3}{2} - \frac{x^6}{2} \]Next, integrate with respect to \( x \):\[ \int_{0}^{1} \left( \frac{x^3}{2} - \frac{x^6}{2} \right) dx = \left[ \frac{x^4}{8} - \frac{x^7}{14} \right]_{0}^{1} \]Evaluate this to get:\[ \frac{1}{8} - \frac{1}{14} = \frac{7 - 4}{56} = \frac{3}{56} \]

Conclusion of the Evaluation

The computed value, \( \frac{3}{56} \), represents the signed volume under the surface \( z = x^2 y \) over the region \( R \). This volume is calculated by integrating in the order specified.

Key concepts

Iterated Integrals

Iterated integrals are a method of evaluating double integrals through a step-by-step process involving nesting one integral inside another. In our exercise, we deal with the region defined by two curves, allowing us to examine the problem through integrals set up in different orders. There's a choice between integrating with respect to \(y\) first, then \(x\), or \(x\) first, then \(y\). Each approach essentially "slices" the area a little differently.

The first configuration, \( \int_{0}^{1} \int_{x^2}^{\sqrt{x}} x^2y \ dy \ dx \), starts by observing how \(y\) changes vertically between the bounds set by the curves \(y = x^2\) (bottom) and \(y = \sqrt{x}\) (top). After solving this level, we integrate in the \(x\) direction from 0 to 1.

The second method, \( \int_{0}^{1} \int_{y^2}^{\sqrt{y}} x^2y \ dx \ dy \), involves first considering the horizontal strips. Here \(x\) varies horizontally from the curve \(x = y^2\) to \(x = \sqrt{y}\), before tackling the \(y\) direction.
These two architectures for setting up the double integral open a versatile approach for tackling the specified region and ensure solution accuracy through flexible integration paths.

Region of Integration

The region of integration is an essential aspect of solving double integrals. In this exercise, the region \(R\) is bounded by the curves \(y = \sqrt{x}\) and \(y = x^2\).

To understand this, envision how these curves behave spatially on a coordinate plane:

• \(y = \sqrt{x}\) forms a curve sweeping upwards from the origin, characteristic of a square root transformation.
• \(y = x^2\) resembles a classic parabola, opening upwards from the origin and expanding slower than the square root curve initially, then faster beyond its intersection point.

The region is the part of the plane where \(y = \sqrt{x}\) is above \(y = x^2\), bounded by their intersection points, \((0,0)\) and \((1,1)\).

It's crucial to correctly sketch and identify these boundaries visually, as any misunderstanding could lead to incorrect integration limits. Our exercise correctly lays out these conditions, setting the scene for seamless computation of volumes or areas.

Intersection Points

Intersection points are critical for defining the region of integration, as they represent the bounds where two curves meet. To find them, we equated \(y = \sqrt{x}\) and \(y = x^2\). Solving this, we set \(\sqrt{x} = x^2\).

This led to solving the equation \(x = x^4\), further breaking down into \(x(x^3 - 1) = 0\), which offers solutions at \(x = 0\) and \(x = 1\). Plugging these into either original equation provides their \(y\)-coordinates, reaffirming their intersections at \((0,0)\) and \((1,1)\).

Knowing the exact intersection points not only confirms the shared boundaries of the region but also guides the limits for our iterated integrals. These precise coordinates are non-negotiable, ensuring the entire area is accurately bounded and computed.

Volume Under Surface

The primary objective of computing a double integral in this context is to find the volume under a surface and above a region in the \(xy\)-plane. In this problem, we're evaluating \(\iint_{R} x^{2} y \, dA\), where \(z = x^2 y\) is the function defining the surface.

This volume represents the accumulation of space between \(z\) and the flat plane (\(z = 0\)) directly over our specified region, \(R\). By setting the iterated integral \(\int_{0}^{1} \int_{x^2}^{\sqrt{x}} x^2y \, dy \, dx\), and performing calculations, we acquire the signed volume beneath the function \(z\).

The result, \(\frac{3}{56}\), is the definitive volume under the curve \(z = x^2 y\) from \(x = 0\) to \(x = 1\) and \(y = x^2\) to \(y = \sqrt{x}\). The term "signed volume" implies consideration of orientation, where all values are summed based on their position relative to the \(xy\)-plane, leading to the most accurate portrayal of the space concerned.