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Chapter 4: Problem 39

Change the order of integration and evaluate the integral. $$ \int_{-1}^{0} \int_{-\sqrt{y+1}}^{\sqrt{y+1}} y^{2} d x d y $$

Short Answer

Expert verified
The value of the integral is \( \frac{16}{15} \).

Step by step solution

01

Understand the Region of Integration

The given integral is \( \int_{-1}^{0} \int_{-\sqrt{y+1}}^{\sqrt{y+1}} y^{2} \ dx \ dy \). This means \( x \) ranges from \( -\sqrt{y+1} \) to \( \sqrt{y+1} \), and \( y \) ranges from \(-1\) to \(0\). This corresponds to a region in the Cartesian plane symmetric about the y-axis and bounded above by the line \(y = 0\) and below by \(y = -1\). The x limits vary with y, forming semicircles.
02

Determine the New Limits of Integration

To switch the order of integration, first consider \( x \) being the outer integral. When \( x \) ranges from \(-1\) to \(1\), \( y \) is bounded between \(-1\) and \(x^2 - 1\) because the upper bound for this region is the semicircle equation \( x = \pm\sqrt{y+1} \), giving \( y = x^2 - 1 \). Hence the new limits are \( -1 \leq x \leq 1 \) and \( -1 \leq y \leq x^2 - 1 \).
03

Rewrite the Integral with New Limits

Rewriting the integral with the new limits gives:\[\int_{-1}^{1} \int_{-1}^{x^2-1} y^2 \ dy \ dx\]
04

Integrate with Respect to y First

Integrate \( y^2 \) with respect to y:\[\int_{-1}^{x^2-1} y^2 \ dy = \left[ \frac{y^3}{3} \right]_{-1}^{x^2-1} = \frac{(x^2-1)^3}{3} - \frac{(-1)^3}{3} = \frac{(x^2-1)^3 + 1}{3}\]
05

Integrate with Respect to x

Now integrate with respect to \( x \):\[\int_{-1}^{1} \left( \frac{(x^2-1)^3 + 1}{3} \right) dx = \frac{1}{3} \left( \int_{-1}^{1} (x^2-1)^3 \ dx + \int_{-1}^{1} 1 \ dx \right)\]
06

Evaluate the Integrals

First calculate \( \int_{-1}^{1} 1 \ dx = [x]_{-1}^{1} = 1 - (-1) = 2 \).Now handle \( \int_{-1}^{1} (x^2-1)^3 \ dx \). Due to symmetry, odd powers of \( x \) over symmetric limits give zero. Compute \( \int_{-1}^{1} (x^6 - 3x^4 + 3x^2 - 1) \ dx \), where the middle terms vanish.The \( x^6 \) term integrates to zero (symmetric, even part), so:\[= \left( 0 - 3 \int_{-1}^{1} x^4 \ dx + 0 - 1(2) \right)\]Integrate \( x^4 \):\[- 3 \left( \frac{x^5}{5} \right)_{-1}^{1} = -3 \left( 2 \frac{1}{5} \right) = -\frac{6}{5}\]Finally, adding all components gives the result:\[\frac{1}{3} \left( \frac{6}{5} + 2 \right) = \frac{1}{3} \cdot \frac{16}{5} = \frac{16}{15}\]
07

Final Step: Compute the Total Integral

The total value of the integral after changing the order and computing is:\[\frac{16}{15}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Change of Integration Order
Changing the order of integration in a double integral can greatly simplify the evaluation process. When dealing with integrals where the limits are variable, like \[ \int_{-1}^{0} \int_{-\sqrt{y+1}}^{\sqrt{y+1}} y^{2} \, dx \, dy \], the process involves re-evaluating the integral by switching the order of integration between the variables. This typically requires redefining the limits of integration by examining the overlap of the regions in different planes.
  • Start by sketching the region of integration to understand the relationship between the variables.
    For the given integral, the region forms a semicircle that is symmetric about the y-axis, ranging from \(y = -1\) to \(y = 0\) and bounded by \(x = \pm \sqrt{y+1}\).
  • To switch the order of integration, redefine constraints so that \(x\) is handled first.
  • Analyze the resulting bounds, which changes the integral limits to \[-1 \leq x \leq 1\] and \(-1 \leq y \leq x^2 - 1\).
Switching the order of integration can often reveal a simpler path to solve the integral, as it uncovers new bounds aligned with more straightforward arithmetic.
Region of Integration
Understanding the region of integration is critical in multi-variable calculus, especially with double integrals. The original double integral \( \int_{-1}^{0} \int_{-\sqrt{y+1}}^{\sqrt{y+1}} y^{2} \, dx \, dy \) represents a specific region in the plane where limits define the shape and size of the area considered.
  • The inner integral's limits \(-\sqrt{y+1} \leq x \leq \sqrt{y+1}\) imply that for each \(y\) value, \(x\) spans the width of a semicircle centered on the y-axis.
  • The outer limits from \(-1\) to \(0\) for \(y\) indicate the vertical extent of this region, creating a bounded segment under the x-axis.
  • This setup segments the Cartesian plane into a wedge-like form that is essential for proper evaluation of the double integral.
  • Correctly interpreting these bounds ensures that calculations adhere to the correct geometric and algebraic properties of the region.
Visualizing the region before integration helps in appropriately setting the variable limits, especially when the order of integration needs changing.
Calculus Integration Techniques
Double integrals are performed by iteratively applying single-variable integration techniques. The trick is to reduce a complex, multi-dimensional problem into manageable, sequential steps. By utilizing different integration methods, we can solve a broader class of problems.
  • First, evaluate the inner integral, treating the outer variable as a constant. This simplifies the task down to evaluating a standard form of single-variable integration.
  • For the given original integral, it begins by integrating \(y^2\) with respect to \(x\) to handle the dependent bounds, resulting in bounds described by \(y+1\).
  • Finally, executing the outer integral involves any techniques suitable for the resulting function, such as polynomial integration, to yield the complete value of the original integral.
  • Here, we used polynomial integration of terms like \((x^2-1)^3\) and evaluated constants:\[\int_{-1}^{1} 1 \, dx \]
These iterative steps ensure precise and comprehensive handling of each dimension involved in the integral, leading toward an accurate final solution.
Iterated Integrals
Iterated integrals are a fundamental concept in calculus used especially for evaluating functions over regions in a plane. Each layer addresses a single dimension of the function's complete behavior.
  • Each integral computes the accumulated sum along a specific axis, with the nested integrals combining these partial results to form a final solution.
  • The process begins by fixing one variable and integrating with respect to another. This considers the contribution of small incremental "slices" or cross-sections across the region.
  • We then substitute these results back into the outer integral for the complete accumulation across the bounding region.
  • In our case, the integration with respect to \(y\) first delivers a simplified expression for subsequent steps, unfurling the complexity into more manageable segments.\(\[ \int_{-1}^{x^2-1} y^2 \, dy \]\)
Through sequential integration and variable order adaptation, iterated integrals provide a robust mechanism for determining the total integral value over a multi-dimensional domain.

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Most popular questions from this chapter

The following problems concern the Theorem of Pappus (see \(\underline{\text { Moments and Centers of Mass for a refresher), a method for calculating }}\) volume using centroids. Assuming a region \(R\), when you revolve around the \(x\) -axis the volume is given by \(V_{x}=2 \pi A \bar{y}\), and when you revolve around the \(y\) -axis the volume is given by \(V_{y}=2 \pi A \bar{x}\), where \(A\) is the area of \(R\). Consider the region bounded by \(x^{2}+y^{2}=1\) and above \(y=x+1\)Find the volume when you revolve the region around the \(y\) -axis.

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