Problem 68
Question
Sketch the region of integration and the solid whose volume is given by the double integral. \begin{equation} \int_{0}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \sqrt{25-x^{2}-y^{2}} d x d y\end{equation}
Step-by-Step Solution
Verified Answer
The integral represents a spherical cap over a semicircle with radius 4.
1Step 1: Understand the Domain of Integration
The given double integral is \( \int_{0}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \sqrt{25-x^{2}-y^{2}} \ dx \ dy \). The limits of integration for \( x \) are from \(-\sqrt{16-y^{2}}\) to \(\sqrt{16-y^{2}}\), which suggests a semicircle in the plane for each fixed \( y \) within a range of a circle \( x^2 + y^2 = 16 \). The outer integral with \( y \) goes from 0 to 4, which indicates the area we are integrating over in the \( y \) direction.
2Step 2: Sketch the Region of Integration
The integrals with respect to \( x \) indicate that for each \( y \) from 0 to 4, \( x \) ranges from \(-\sqrt{16-y^{2}}\) to \(\sqrt{16-y^{2}}\). This implies that the region of integration is the area under a semicircle of radius 4, centered at the origin, and restricted to the half where \( y \) is positive (0 to 4). Hence, the region is a semicircle in the upper half of the circle \( x^2 + y^2 = 16 \).
3Step 3: Understand the Function of Integration
The function \( \sqrt{25-x^{2}-y^{2}} \) represents the upper surface of the solid. This can be recognized as the top half of a sphere's equation \( x^2 + y^2 + z^2 = 25 \). This indicates that above each semicircular slice of our integration region, we have a spherical cap whose highest point is at \( z=5 \), when \( x = 0, y = 0 \).
4Step 4: Sketch the Solid Whose Volume is Represented
The solid whose volume is represented by the integral is a spherical cap. This cap is bounded by the half-sphere defined by \( x^2 + y^2 + z^2 = 25 \) on the range where \( x^2 + y^2 \leq 16 \), and \( y \) from 0 to 4. This forms a shape wherein the base is a semicircle of radius 4 on the \( xy \)-plane, and the top is a curved surface of the sphere.
Key Concepts
Understanding the Region of IntegrationAnalyzing the Spherical CapDecoding the Semicircle AreaUnderstanding Volume Calculation
Understanding the Region of Integration
A double integral involves integrating over a specific region within a plane. In this case, the double integral is expressed as \( \int_{0}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \sqrt{25-x^{2}-y^{2}} \, dx \, dy \).
Here, the limits for \( x \) suggest that for each specific value of \( y \), \( x \) ranges from \( -\sqrt{16-y^{2}} \) to \( \sqrt{16-y^{2}} \), forming a semicircle. The upper and lower limits of the outer integral with respect to \( y \), from 0 to 4, indicate the vertical section within which this semicircle lies.
In simpler terms, as \( y \) varies from 0 to 4, we gather all \( x \) values that can form a semicircle, which reflects a part of the circle \( x^2 + y^2 = 16 \). This means the region of integration stretches over the
Here, the limits for \( x \) suggest that for each specific value of \( y \), \( x \) ranges from \( -\sqrt{16-y^{2}} \) to \( \sqrt{16-y^{2}} \), forming a semicircle. The upper and lower limits of the outer integral with respect to \( y \), from 0 to 4, indicate the vertical section within which this semicircle lies.
In simpler terms, as \( y \) varies from 0 to 4, we gather all \( x \) values that can form a semicircle, which reflects a part of the circle \( x^2 + y^2 = 16 \). This means the region of integration stretches over the
- Upper half of a circle with radius 4
- Symmetrical across the x-axis with \( y \) values limited from 0 to 4
Analyzing the Spherical Cap
The function we are integrating is \( \sqrt{25-x^{2}-y^{2}} \), representing the upper hemisphere of a sphere with a complete equation \( x^2 + y^2 + z^2 = 25 \).
This function, as a result, forms a spherical cap—a portion of a sphere cut off by a plane.
At any chosen point \((x, y)\) in the defined region, this tells us the height \( z \) extends up to \( \sqrt{25-x^2-y^2} \). This correlation yields the familiar shape of a dome capped from the top of the sphere.
The spherical cap thus:
This function, as a result, forms a spherical cap—a portion of a sphere cut off by a plane.
At any chosen point \((x, y)\) in the defined region, this tells us the height \( z \) extends up to \( \sqrt{25-x^2-y^2} \). This correlation yields the familiar shape of a dome capped from the top of the sphere.
The spherical cap thus:
- Is built over the semicircle identified in the \( xy \)-plane.
- Rises to a maximum height of 5 units when \( x=0 \) and \( y=0 \), reaching the top point of this hemisphere.
Decoding the Semicircle Area
The semicircle forms the base of our spherical cap in the integration process. To visualize this part, think of a circle \( x^2 + y^2 = 16 \), which is cut in half along the x-axis.
Even though the equation delineates a full circle, by setting the restrictions \( y \) only between 0 and 4, we effectively consider just the upper hemisphere.
This semicircle:
Even though the equation delineates a full circle, by setting the restrictions \( y \) only between 0 and 4, we effectively consider just the upper hemisphere.
This semicircle:
- Has a radius of 4—that’s because \( x^2 + y^2 \leq 16 \) outlines a full circle with this radius.
- Occupies space only where \( y \) is positive from 0 to 4.
Understanding Volume Calculation
Volume calculation in the context of a double integral involves assessing the space under a surface across the region of integration. The double integral \( \int_{0}^{4} \int_{-\sqrt{16-y^{2}}}^{\sqrt{16-y^{2}}} \sqrt{25-x^{2}-y^{2}} \, dx \, dy \) essentially computes the volume of the spherical cap.
This is achieved by evaluating the sum of all infinitesimally small volumes \( \sqrt{25-x^2-y^2} \, dx \, dy \) over the integration region. Since these define the height as \( z \) above every point \( (x, y) \), the computation extends upward within the hemisphere.
When implemented:
This is achieved by evaluating the sum of all infinitesimally small volumes \( \sqrt{25-x^2-y^2} \, dx \, dy \) over the integration region. Since these define the height as \( z \) above every point \( (x, y) \), the computation extends upward within the hemisphere.
When implemented:
- The inner integral calculates the width along the x-axis for each slice of y.
- The outer integral sums up these slices as y moves from 0 to 4.
Other exercises in this chapter
Problem 66
Find the volume of the solid bounded on the front and back by the planes \(x=\pm \pi / 3,\) on the sides by the cylinders \(y=\pm \sec x,\) above by the cylinde
View solution Problem 67
In Exercises 67 and \(68,\) sketch the region of integration and the solid whose volume is given by the double integral. \begin{equation}\int_{0}^{3} \int_{0}^{
View solution Problem 71
Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is cond
View solution Problem 71
Sphere and cones Find the volume of the portion of the solid sphere \(\rho \leq a\) that lies between the cones \(\phi=\pi / 3\) and \(\phi=2 \pi / 3\)
View solution