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Chapter 15: Problem 60

Find the volume of the region bounded above by the paraboloid \(z=9-x^{2}-y^{2},\) below by the \(x y\) -plane, and lying outside the cylinder \(x^{2}+y^{2}=1\)

Short Answer

Expert verified
The volume is \(32\pi\).

Step by step solution

01

Identify the Region for Integration

The region is bounded above by the paraboloid \(z = 9 - x^2 - y^2\) and below by the xy-plane. Additionally, the region lies outside the cylinder \(x^2 + y^2 = 1\), which is a circle in the xy-plane with radius 1. We need to set up a triple integral in cylindrical coordinates, as the problem is symmetrical around the z-axis.
02

Convert to Cylindrical Coordinates

In cylindrical coordinates, \(x = r\cos\theta\), \(y = r\sin\theta\), and \(z = z\). The equation of the paraboloid becomes \(z = 9 - r^2\). The volume lies above the region \(r > 1\) because it is outside the cylinder \(x^2 + y^2 = 1\).
03

Set up the Limits of Integration

For \(\theta\), it ranges from 0 to \(2\pi\) due to the full rotation around the z-axis. For \(r\), it ranges from 1 to \(3\) because \(z = 0\) implies \(r^2 = 9\) so \(r = 3\). For \(z\), it ranges from 0 to \(9 - r^2\).
04

Write the Triple Integral for Volume

The volume \(V\) can be expressed as:\[ V = \int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{9-r^2} r \, dz \, dr \, d\theta \]
05

Integrate with Respect to z

Integrate the innermost integral:\[ \int_{0}^{9-r^2} r \, dz = r[z]_{0}^{9-r^2} = r(9-r^2) \]
06

Integrate with Respect to r

Substitute the result back into the double integral and integrate over \(r\):\[ \int_{1}^{3} r(9-r^2) \, dr = \int_{1}^{3} (9r - r^3) \, dr \]\[ = \left[ \frac{9}{2}r^2 - \frac{r^4}{4} \right]_{1}^{3} = \left(\frac{9}{2}(9) - \frac{81}{4}\right) - \left(\frac{9}{2}(1) - \frac{1}{4}\right) \]\[ = \left(\frac{81}{2} - \frac{81}{4}\right) - \left(\frac{9}{2} - \frac{1}{4}\right) \]
07

Simplify the Integration Result

Simplifying the expression for the integral over \(r\):\[ = \frac{162}{4} - \frac{81}{4} - \frac{18}{4} + \frac{1}{4} = \frac{162 - 81 - 18 + 1}{4} = \frac{64}{4} = 16 \]
08

Integrate with Respect to \( \theta \)

The volume integral now becomes:\[ V = \int_{0}^{2\pi} 16 \, d\theta = 16 \theta \left|_{0}^{2\pi} \right. = 16 \times 2\pi = 32\pi \]
09

Final Step: Conclusion

The volume of the region bounded by the paraboloid and outside the cylinder is \(32\pi\).

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Key Concepts

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

Cylindrical Coordinates
When solving problems involving volumes of solids, using cylindrical coordinates can be highly effective. Cylindrical coordinates are particularly useful when dealing with symmetrical shapes around the z-axis, like cylinders and paraboloids. In this system, any point in space is expressed using three values:
  • \( r \) 鈥� the radial distance from the z-axis
  • \( \theta \) 鈥� the angle between the point and the positive x-axis, measured in radians
  • \( z \) 鈥� the height above the xy-plane
The transformation from Cartesian coordinates \((x, y, z)\) to cylindrical coordinates \((r, \theta, z)\) is given by:
  • \( x = r\cos\theta \)
  • \( y = r\sin\theta \)
  • \( z = z \)
Using these, the problem of calculating the volume of a solid becomes easier when dealing with rotational symmetry.
Triple Integral
A triple integral is employed to calculate volumes, among other things, in three-dimensional space. It enables us to account for variations along all three spatial dimensions - in this case, through the variables \(r\), \(\theta\), and \(z\). When setting up a triple integral in cylindrical coordinates, the integration is typically performed over:
  • \( \theta \) from 0 to \(2\pi\), to complete a full circle around the z-axis
  • \( r \) from a specified lower to upper limit, defining the radial extent
  • \( z \) from its lower bound (often the xy-plane where \(z=0\)) up to a function indicating its upper extent
In this exercise, the volume of the solid is expressed using: \[ \int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{9-r^2} r \, dz \, dr \, d\theta \]The innermost integral represents the integration in the vertical direction, followed by radial integration, and ending with the theta integration to account for rotation.
Paraboloid
A paraboloid is a three-dimensional object that looks like a parabolic dish, defined by a quadratic equation. In this problem, the paraboloid is given by the equation:\[ z = 9 - x^2 - y^2 \] This shape opens downward, meaning it curves downward as you move away from the center in the xy-plane. The highest point on this paraboloid is at \( z = 9 \) when both \( x \) and \( y \) are zero.In cylindrical coordinates, this equation becomes:\[ z = 9 - r^2 \]This representation shows that the height \( z \) is dependent on \( r \), the distance from the origin in the xy-plane. The boundary for \( z \) thus decreases quadratically as \( r \) increases.
Cylinder
In the problem, the cylinder is defined by the equation:\[ x^2 + y^2 = 1 \] This represents a vertical cylinder centered on the z-axis with a radius of 1. In cylindrical coordinates, transforming the variables gives a simple expression:\[ r^2 = 1 \]This shows that the radial coordinate \( r \) is fixed at 1. Thus, the region of interest in the exercise is outside this cylinder for radii \( r > 1 \). The cylinder acts as an inner boundary, and the region's volume is constrained outside this space while being bounded above by the paraboloid.

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Most popular questions from this chapter

Give the limits of integration for evaluating the integral $$\iiint f(r, \theta, z) d z r d r d \theta$$ as an iterated integral over the region that is bounded below by the plane \(z=0,\) on the side by the cylinder \(r=\cos \theta,\) and on top by the paraboloid \(z=3 r^{2}\)

Gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. $$\left.\int_{0}^{3 / 2} \int_{1}^{4-2 u} \frac{4-2 u}{v^{2}} d v d u \quad \text { (the } u v \text { -plane }\right)$$

fInd a. the mass of the solid. b. the center of mass. c. the moments of inertia about the coordinate axes. A solid cube in the first octant is bounded by the coordinate planes and by the planes \(x=1, y=1,\) and \(z=1 .\) The density of the cube is \(\delta(x, y, z)=x+y+z+1.\)

Approximate the double integral of \(f(x, y)\) over the region \(R\) partitioned by the given vertical lines \(x=a\) and horizontal lines \(y=c .\) In each subrectangle, use \(\left(x_{k}, y_{k}\right)\) as indicated for your approximation. $$\iint_{R} f(x, y) d A=\sum_{i=1}^{n} f\left(x_{k}, y_{k}\right) \Delta A_{k}$$ \(f(x, y)=x+2 y\) over the region \(R\) inside the circle \((x-2)^{2}+(y-3)^{2}=1\) using the partition \(x=1,3 / 2,2,5 / 2\) 3 and \(y=2,5 / 2,3,7 / 2,4\) with \(\left(x_{k}, y_{k}\right)\) the center (centroid) in the \(k\) th subrectangle (provided the subrectangle lies within \(R\) )

Sketch the region of integration, reverse the order of integration, and evaluate the integral. $$\int_{0}^{1 / 16} \int_{y^{1 / 2}}^{1 / 2} \cos \left(16 \pi x^{5}\right) d x d y$$
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