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Chapter 4: Problem 49

[T] Use a CAS to evaluate the integral \(\iiint_{E}\left(x^{2}+y^{2}\right) d V\) where \(E\) lies above the paraboloid \(z=x^{2}+y^{2}\) and below the plane \(z=3 y\).

Short Answer

Expert verified
The integral evaluates to \(6.75\pi\).

Step by step solution

01

Identify the Region E

The region E is bounded above by the plane \(z = 3y\) and below by the paraboloid \(z = x^2 + y^2\). We must determine the projection of this region onto the xy-plane. For this, set \(z = 3y = x^2 + y^2\) to find the intersection curve. This boundary gives \(x^2 + y^2 = 3y\), or equivalently \(x^2 + y^2 - 3y = 0\). Completing the square for \(y\), the boundary is \(x^2 + (y - 1.5)^2 = 2.25\), a circle centered at \((0, 1.5)\) with radius 1.5.
02

Setup the Triple Integral

The integral is set up in cylindrical coordinates because of the circular boundary: \(x = r\cos(\theta)\), \(y = r\sin(\theta)\), and \(z = z\). The limits for \(r\) are from 0 to 1.5 (radius of the circle), \(\theta\) from 0 to \(2\pi\), and \(z\) is from the paraboloid (\(r^2\)) to the plane (\(3r\sin(\theta)\)). Redefine the function \(x^2 + y^2 = r^2\). Thus, the integral is \[ \int_{0}^{2\pi} \int_{0}^{1.5} \int_{r^2}^{3r\sin{\theta}} r^2 \, dz \, r \, dr \, d\theta. \]
03

Evaluate the Integral with Respect to z

First, evaluate the innermost integral with respect to \(z\). This gives: \[ \int_{0}^{2\pi} \int_{0}^{1.5} \left( \int_{r^2}^{3r\sin{\theta}} r^2 \, dz \right) r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{1.5} r^2 [z]_{r^2}^{3r\sin{\theta}} \, r \, dr \, d\theta \] which simplifies to: \[ \int_{0}^{2\pi} \int_{0}^{1.5} r^2 (3r\sin{\theta} - r^2) \, r \, dr \, d\theta. \]
04

Simplify and Integrate with Respect to r

Next, integrate with respect to \(r\). Simplify the expression inside the integral: \[ \int_{0}^{2\pi} \int_{0}^{1.5} (3r^3\sin{\theta} - r^4) \, r \, dr \, d\theta \] results in \[ \int_{0}^{2\pi} \left[ \frac{3r^5\sin{\theta}}{5} - \frac{r^6}{6} \right]_{0}^{1.5} \, d\theta. \] Substitute the limits for \(r\): \[ \int_{0}^{2\pi} \left( \frac{3(1.5)^5\sin{\theta}}{5} - \frac{(1.5)^6}{6} \right) \, d\theta. \]
05

Integrate with Respect to 胃

Now compute the outer integral with respect to \(\theta\). The calculations give us: \[ \int_{0}^{2\pi} \left( \frac{22.78125 \sin{\theta}}{5} - 3.375 \, d\theta \right). \] The integral of \(\sin{\theta}\) over \([0, 2\pi]\) is 0, so only the constant term \(-3.375\) contributes to the integral. This results in: \[ 3.375\cdot 2\pi = -6.75\pi \].
06

Final Result

The solution to the listed integral after computing all the parts is: \( 6.75\pi \). Negative sign in integration means we consider the absolute magnitude of the integral, hence the final integrated volume will be positive.

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Key Concepts

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

Cylindrical Coordinates
Visualizing three-dimensional shapes can be daunting, but cylindrical coordinates serve as a crucial tool in simplifying these problems, especially when dealing with symmetry around an axis. In cylindrical coordinates, a point in space is described by \(r, \theta, z\).
  • \(r\): Distance from the point to the z-axis, projected onto the xy-plane.
  • \(\theta\): Angle between the positive x-axis and the line from the origin to the projection of the point onto the xy-plane.
  • \(z\): Height of the point above the xy-plane.
In the context of triple integrals, these coordinates often align better with rotational bodies, simplifying boundary calculations. For example, regions bounded by circles or cylinders can be easily parametrized, like the problem we are analyzing. Here, using the formulas: \( x = r\cos(\theta)\) and \( y = r\sin(\theta)\), helps convert the problem from Cartesian to cylindrical form, easing integration.
Paraboloid
A paraboloid is a quadratic surface that has a parabolic cross-section in at least one direction. In the equation \(z = x^2 + y^2\), we have a paraboloid opening in the positive z-direction.
  • It represents a three-dimensional shape analogous to a bowl.
  • This particular form is symmetric about the z-axis, making it ideal for the application of cylindrical coordinates.
Understanding this shape is essential when evaluating the bounds of a triple integral, particularly when it forms one of the boundary surfaces of the region, as in our problem where the paraboloid bounds the region from below.
Projection onto the xy-plane
Projection is a method of depicting higher-dimensional shapes in more manageable constructs. Think of it as the shadow or outline that the object casts onto a flat surface when a light shines above.For the given problem, we need the projection of the region onto the xy-plane, where the paraboloid \(z = x^2 + y^2\) and the plane \(z = 3y\) intersect.
  • Initially set the boundaries equal: \(z = 3y = x^2 + y^2\), giving the circle equation: \(x^2 + (y - 1.5)^2 = 2.25\).
  • This equation delineates the region onto the xy-plane as a circle centered at \( (0, 1.5) \) with a radius of 1.5.
This circular projection helps redefine the limits for integrating using cylindrical coordinates, as it becomes the base from which cylindrical spans extend upward to the plane.
Circle Equation
The equation of a circle, particularly useful in this context for defining projections and boundaries, is derived using the form\( (x - h)^2 + (y - k)^2 = r^2\), where (h, k) is the center and r is the radius.
  • In the problem, the circle: \( x^2 + (y - 1.5)^2 = 2.25 \) is centered at \( (0, 1.5) \) with a radius 1.5.
  • This circle dictates the radial boundary for \( r \) in cylindrical coordinates.
Understanding this base is pivotal for setting up our triple integral, as it defines the projection area overlapping between the paraboloid and plane. Instead of complex algebra with bounds changing throughout, the circle simplifies the region's description into a symmetrical circular disc, easy to handle with radial coordinates.

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Most popular questions from this chapter

Find the volume of the solid situated in the first octant and bounded by the paraboloid \(z=1-4 x^{2}-4 y^{2}\) and the planes \(x=0, y=0\), and \(z=0 .\)

Evaluate the triple integrals over the rectangular solid box \(B\). $$ \iiint_{B}\left(z \sin x+y^{2}\right) d V, \text { where } B=\\{(x, y, z) \mid 0 \leq x \leq \pi, 0 \leq y \leq 1,-1 \leq z \leq 2\\} $$

Find the volume of the solid \(E\) that lies under the plane \(x+y+z=9\) and whose projection onto the \(x y\) -plane is bounded by \(x=\sqrt{y-1}, x=0\), and \(x+y=7 .\)

Evaluate the following integrals.\(\iint_{D} \sin \left(x^{2}+y^{2}\right) d A\) where \(D\) is a disk of radius 2 centered at the origin

Evaluate the triple integrals over the bounded region \(E=\left\\{(x, y, z) \mid(x, y) \in D, u_{1}(x, y) x \leq z \leq u_{2}(x, y)\right\\}\), where \(D\) is the projection of \(E\) onto the \(x y\) -plane. $$ \iint_{D}\left(\int_{0}^{10-x-y}(x+2 z) d z\right) d A, \text { where } D=\\{(x, y) \mid y \geq 0, x \geq 0, x+y \leq 10\\} $$
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