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Chapter 13: Problem 20

Sketch the solid whose volume is given by the following double integrals over the rectangle \(R=\\{(x, y)\) : \(0 \leq x \leq 2,0 \leq y \leq 3\\}\) $$ \iint_{R}\left(25-x^{2}-y^{2}\right) d A $$

Short Answer

Expert verified
It's a portion of a downward-opening paraboloid over the rectangle \( R \).

Step by step solution

01

Understand the Given Integral

You are given a double integral over a region \( R \), represented by the rectangle \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 3 \). The integrand is \( 25 - x^2 - y^2 \), which suggests a parabolic shape.
02

Identify the Function

The function \( f(x, y) = 25 - x^2 - y^2 \) is a downward-opening paraboloid. This type of function forms a 3D surface where the height \( z \) decreases as \( x \) or \( y \) increases.
03

Analyze the Domain

The integration is over the rectangle \( R \) in the \( xy \)-plane. For points within the rectangle, the values of \( x \) range from 0 to 2, and \( y \) ranges from 0 to 3. This defines the bounds for our sketch.
04

Understand the Boundaries of the Shape

At the boundaries of the rectangular domain, evaluate \( f(x, y) \):- When \( x = 0 \) and \( y = 0 \), \( z = 25 \).- When \( x = 0 \) and \( y = 3 \), \( z = 16 \).- When \( x = 2 \) and \( y = 0 \), \( z = 21 \).- When \( x = 2 \) and \( y = 3 \), \( z = 12 \).These values give us key points to outline the 3D shape.
05

Sketch the Solid

Sketch the paraboloid on the rectangle defined by \( R \) in the xy-plane. The paraboloid opens downwards from a peak of \( z = 25 \). Varying heights at boundaries indicate the surface slopes downwards as you move away from the origin within the rectangle.
06

Complete Visualization

The solid is a portion of the paraboloid capped at the base by the rectangle \( R \). Visualize it as a 3D surface starting high at the origin and sloping down as \( x \) and \( y \) reach their maximum bounds. Complete the sketch with curves to indicate the paraboloid's surface.

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Key Concepts

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

Paraboloid
A paraboloid is a three-dimensional shape that looks like a bowl or a dish. Specifically, the paraboloid in our exercise is described by the function \( f(x, y) = 25 - x^2 - y^2 \). Here, it is a downward-opening paraboloid. This means it starts at its highest point and slopes downwards.
  • The highest point, or peak, of this paraboloid is at \( (0, 0, 25) \), where the value 25 is the maximum height.
  • As \( x \) or \( y \) increase, the value of \( z \) (height) decreases because \( x^2 \) and \( y^2 \) are subtracted from 25.
This shape is important because it represents the surface above which we want to calculate the volume. By understanding the paraboloid, you can better visualize the 3D shape involved. The paraboloid's axis is aligned with the z-axis, creating a symmetrical shape around it.
Volume Calculation
Volume calculation using double integrals involves summing small volumes above each small area within a specified region. In our case, the region is defined by the rectangle \( R \) with bounds: \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 3 \).
  • Each small volume element is given by \( f(x, y) \cdot dA \), where \( f(x, y) = 25 - x^2 - y^2 \)
  • \( dA \) represents a tiny area in the \( xy \)-plane, which when multiplied by the height \( f(x, y) \), gives a small volume.
  • We use a double integral \( \iint_{R} (25 - x^2 - y^2) \, dA \) to sum these elements over the entire region \( R \).
The value of the double integral gives us the total volume of the solid under the paraboloid and above the rectangle. Completing this integral involves evaluating the integrand over the region's bounds, adding all those little slices together to get the whole volume.
3D Sketching
Creating a sketch of the solid is a crucial step in visualizing the three-dimensional structure of the paraboloid over the region \( R \). This visualization helps in understanding both the boundaries and the shape of the solid. Follow these steps to make an accurate sketch:
  • Start by drawing the rectangle \( R \) on the \( xy \)-plane: it stretches from \( x = 0 \) to \( x = 2 \) and \( y = 0 \) to \( y = 3 \).
  • Mark the key boundary values for \( z \): this includes points like \((0, 0, 25)\), \((0, 3, 16)\), \((2, 0, 21)\), and \((2, 3, 12)\).
  • Begin sketching the paraboloid starting high at the origin (\(0,0\)) and sloping downwards as \(x\) and \(y\) vary within the rectangle.
As you draw, imagine the surface of the paraboloid as a curved slope, smoothly descending to its edges at the boundary. Curved lines represent the slope, giving the figure depth and a three-dimensional perspective. This helps you better grasp the physical aspect of the problem you are solving.

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Most popular questions from this chapter

Evaluate the iterated integrals. $$ \int_{1}^{2} \int_{0}^{x-1} y d y d x $$

Evaluate the given double integral by changing it to an iterated integral. $$ \begin{aligned} &\text { 16. } \iint_{S}(x+y) d A ; S \text { is the triangular region with vertices }\\\ &(0,0),(0,4), \text { and }(1,4) \end{aligned} $$.

A full soda can of height \(h\) stands on the \(x y\) -plane. Punch a hole in the base and watch \(\bar{z}\) (the \(z\) coordinate of the center of mass) as the soda leaks away. Starting at \(h / 2, \bar{z}\) gradually drops to a minimum and then rises back to \(h / 2\) when the can is empty. Show that \(\bar{z}\) is least when it coincides with the height of the soda. (Do not neglect the mass of the can itself.) Would the same conclusion hold for a soda bottle? Hint: Don't calculate; think geometrically.

Write the given iterated integral as an iterated integral with the indicated order of integration. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} \int_{0}^{\sqrt{1-y^{2}-z^{2}}} f(x, y, z) d x d z d y ; d z d y d x $$

Write the given iterated integral as an iterated integral with the order of integration interchanged. Hint: Begin by sketching a region \(S\) and representing it in two ways. $$ \int_{0}^{1} \int_{0}^{x} f(x, y) d y d x $$
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