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Chapter 15: Problem 13

Evaluate the integrals. $$ \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \int_{0}^{\sqrt{9-x^{2}}} d z d y d x $$

Short Answer

Expert verified
The answer is 18.

Step by step solution

01

Understand the Triple Integral

The given problem is a triple integral in the form \[ \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \int_{0}^{\sqrt{9-x^{2}}} dz \, dy \, dx \].This represents the volume under the surface defined by \(z = \sqrt{9-x^2}\) within the bounds provided for \(x\), \(y\), and \(z\).
02

Simplify the Integral

Since the inner integral \( \int_{0}^{\sqrt{9-x^{2}}} dz \) is with respect to \(z\), we can compute this first. The integral is from \( 0 \) to \( \sqrt{9 - x^2} \), so it simplifies to \( \sqrt{9-x^2} \). The integral evaluates to \[ \left[z\right]_0^{\sqrt{9-x^2}} = \sqrt{9-x^2} - 0 = \sqrt{9-x^2} \].
03

Evaluate Second Integral with Respect to y

Substituting the result of the first integral into the second integral, we have: \[ \int_{0}^{\sqrt{9-x^2}} \sqrt{9-x^2} \, dy \].Since \( \sqrt{9-x^2} \) is a constant with respect to \( y \), this simplifies to:\[ \sqrt{9-x^2} \cdot y \bigg|_0^{\sqrt{9-x^2}} = \sqrt{9-x^2} \cdot \sqrt{9-x^2} = 9-x^2 \].
04

Evaluate the Outer Integral

We now evaluate the integral with respect to \(x\):\[ \int_{0}^{3} (9-x^2) \, dx \].Split this into two separate integrals:\[ \int_{0}^{3} 9 \, dx - \int_{0}^{3} x^2 \, dx \].The first integral evaluates to \[ 9x \bigg|_0^3 = 27 \].The second integral evaluates to \[ \frac{x^3}{3} \bigg|_0^3 = 9 \].Subtracting these gives \[ 27 - 9 = 18 \].
05

Write the Final Answer

The result of evaluating the triple integral is \( 18 \). This is the volume under the surface defined by the specified region in the triple integral.

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Key Concepts

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

Integration Techniques
Triple integrals are a powerful integration method used in multivariable calculus. They consist of nested integrals, where each integral is evaluated over a specific variable within a given boundary.
In this exercise, we navigate through the nested integrals starting from the innermost to the outermost.
  • First, identify the order of integration, which in our problem is \(dz \, dy \, dx\).
  • The innermost integral, \(\int_{0}^{\sqrt{9-x^{2}}} dz\), is integral with respect to z. It evaluates the z-bounds from 0 to \sqrt{9-x^2}. \
  • This results in simplification into the term \( \sqrt{9-x^2} \, \times \, \text{dy dx} \).
This process of step-by-step simplification ensures precision and understanding at each stage.
Breaking down the integral in a systematic manner with respect to each variable is vital for accurate results in volume calculations.
Volume Calculation
In multivariable calculus, triple integrals are frequently used for volume calculation of three-dimensional objects. The given exercise focuses on finding the volume under a curve defined by specific bounds.
The goal is to account for the accumulated volume in small slice-by-slice fashion:
  • After evaluating each integral, replace the solved expression back into the main equation to continue building upon the previous results.
  • The inner two integrals combine to cover a volume segment defined by \( \sqrt{9-x^2} \) times the area defined by \( y \) limits.
  • The final integration step accounts for the entire x-boundary, creating a solid three-dimensional volume result.
This layered approach ensures all spatial dimensions are integrated, offering a complete picture of the bounded volume.
Multivariable Calculus
Multivariable calculus expands the concept of calculus beyond single-variable functions. Triple integrals extend integration into three dimensions, assessing variation across more than one input.
  • Mainly used to find volumes, but also applicable in diverse fields such as physics and engineering, where multiple variables define a scenario.
  • In these calculus problems, it's essential to understand how each variable interaction influences the resulting geometric object.
  • The understanding of coordinate systems and set limits are crucial for correctly setting problems, especially when working with Cartesian or polar coordinates.
By tackling this triple integral, students engage with critical thinking and problem-solving skills necessary for mastering these complex computations in multivariable environments.

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Most popular questions from this chapter

The Parallel Axis Theorem Let \(L_{\mathrm{cm}}\) be a line through the center of mass of a body of mass \(m\) and let \(L\) be a parallel line \(h\) units away from \(L_{\mathrm{cm}} .\) The Parallel Axis Theorem says the moments of inertia \(I_{\mathrm{cm}}\) and \(I_{L}\) of the body about \(L_{\mathrm{cm}}\) and \(L\) satisfy the equation $$I_{L}=I_{\mathrm{c.m.}}+m h^{2}$$ As in the two-dimensional case, the theorem gives a quick way to calculate one moment when the other moment and the mass are known.\ Proof of the Parallel Axis Theorem a. Show that the first moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the \(y z\) -plane. What does the formula \(\overline{x}=M_{y z} / M\) then tell you? b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line \(L_{c . m .}\) along the \(z\) -axis and the line \(L\) perpendicular to the \(x y\) -plane at the point \((h, 0,0)\) . Let \(D\) be the region of space occupied by the body. Then, in the notation of the figure, $$I_{L}=\iiint_{D}|\mathbf{v}-h \mathbf{i}|^{2} d m$$ Expand the integrand in this integral and complete the proof.

Evaluate the integral $$ \int_{0}^{4} \int_{x=y / 2}^{x=(y / 2)+1} \frac{2 x-y}{2} d x d y $$ from Example 1 directly by integration with respect to \(x\) and \(y\) to confirm that its value is 2 .

Use a CAS integration utility to evaluate the triple integral of the given function over the specified solid region. \(F(x, y, z)=x^{4}+y^{2}+z^{2}\) over the solid sphere \(x^{2}+y^{2}+\) \(z^{2} \leq 1\)

Centroid of a solid semiellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry three-eighths of the way from the base toward the top, show, by transforming the appropriate integrals, that the center of mass of a solid semiellipsoid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)+\left(z^{2} / c^{2}\right) \leq 1, \quad z \geq 0\) lies on the \(z\) -axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals. $$ \int_{\pi / 6}^{\pi / 2} \int_{-\pi / 2}^{\pi / 2} \int_{\csc \phi}^{2} 5 \rho^{4} \sin ^{3} \phi d \rho d \theta d \phi $$
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