Problem 30
Question
The region in the first octant bounded by the coordinate planes and the surface \(z=4-x^{2}-y\) (GRAPH NOT COPY)
Step-by-Step Solution
Verified Answer
The region is bounded by \(x^2 + y = 4\), forming a parabola in the first octant.
1Step 1: Understand the Surface Equation
The given equation is \(z = 4 - x^2 - y\). This represents a surface in the 3D space. Since we are dealing with the first octant, \(x\), \(y\), and \(z\) are all non-negative.
2Step 2: Determine the Intersection with the Coordinate Planes
The first octant is constrained by \(x = 0\), \(y = 0\), and \(z = 0\). We need to find the intersections of these planes with the given surface:- When \(z = 0\), substituting into the equation gives \(0 = 4 - x^2 - y\) or \(x^2 + y = 4\), which is a parabola in the \(xy\)-plane.- When \(x = 0\), we have \(z = 4 - y\), intersecting the \(yz\)-plane, an equation of a line. - When \(y = 0\), we have \(z = 4 - x^2\), intersecting with the \(xz\)-plane, which is also a parabola.
3Step 3: Find Limits for x, y, and z
From the equation \(x^2 + y = 4\), solve for \(y\): \(y = 4 - x^2\). Maximum \(y\) is 4 when \(x = 0\) and minimum \(y\) is 0 when \(x^2 = 4\) or \(x = 2\), making \(x\) range from 0 to 2.
4Step 4: Sketch or Visualize the Shaped Region
The shape formed in the first octant is bounded by three planes and the curve \(x^2 + y = 4\), forming a 3D volume underneath the surface of \(z = 4 - x^2 - y\).
5Step 5: Set Up Integral for Volume (Optional)
If the task requires finding the volume of the region, set up a triple integral with the bounds:\[\int_0^2 \int_0^{4-x^2} \int_0^{4-x^2-y} \, dz \, dy \, dx.\]This integral calculates the total volume within the bounds of the equations determined.
Key Concepts
Triple IntegralsRegions in SpaceSurfaces and Planes
Triple Integrals
Triple integrals are a powerful tool in multivariable calculus, allowing us to calculate volumes of three-dimensional regions. Imagine slicing a 3D shape into many tiny rectangular prisms. A triple integral sums up the volumes of all these tiny prisms. This sum gives us the total volume of the 3D shape.
When setting up a triple integral, you will often encounter the notation \[ \int \int \int \]. The limits of these integrals define the region over which you are integrating. For instance, in our exercise, the integral is \[ \int_0^2 \int_0^{4-x^2} \int_0^{4-x^2-y} \, dz \, dy \, dx \]. Each pair of limits corresponds to one of the three variables \(x\), \(y\), and \(z\).
When setting up a triple integral, you will often encounter the notation \[ \int \int \int \]. The limits of these integrals define the region over which you are integrating. For instance, in our exercise, the integral is \[ \int_0^2 \int_0^{4-x^2} \int_0^{4-x^2-y} \, dz \, dy \, dx \]. Each pair of limits corresponds to one of the three variables \(x\), \(y\), and \(z\).
- The innermost integral, with respect to \(z\), accumulates the small slices along the \(z\)-axis, given the values of \(x\) and \(y\).
- The middle integral, with respect to \(y\), sums these slices as \(y\) varies.
- Finally, the outermost integral, with respect to \(x\), combines all these contributions over the range of \(x\).
Regions in Space
Regions in space are specific areas within three-dimensional space that are bounded by surfaces and planes. In our example, we focus on the region in the first octant. The first octant is where all three coordinates \(x\), \(y\), and \(z\) are positive.
Defining this region involves understanding the constraints posed by the surrounding surfaces and planes. The key idea here is that the region enclosed should satisfy all equations or inequalities simultaneously.
Defining this region involves understanding the constraints posed by the surrounding surfaces and planes. The key idea here is that the region enclosed should satisfy all equations or inequalities simultaneously.
- The plane \(z = 0\) denotes the bottom of the region, where \(z\) scores start, and it's the base of our volume.
- The equation \(z = 4 - x^2 - y\) creates a curved cap over the region, shaping it based on the values \(x\) and \(y\).
- Finally, the planes \(x = 0\) and \(y = 0\) mark the sides of the volume, restricting our calculations to the first octant.
Surfaces and Planes
Surfaces and planes are fundamental geometric elements in multivariable calculus. A surface in 3D space is a two-dimensional shape extending infinitely or bounded by certain conditions, whereas a plane is typically a flat, infinite surface.
In our exercise example, the surface is defined by the equation \(z = 4 - x^2 - y\). This surface forms a paraboloid, a shape that smoothly curves downward from the top vertex at \((0,0,4)\).
In our exercise example, the surface is defined by the equation \(z = 4 - x^2 - y\). This surface forms a paraboloid, a shape that smoothly curves downward from the top vertex at \((0,0,4)\).
- Here, \(z\) decreases as either \(x\) or \(y\) increases, resulting in a downward-opening bowl shape.
- Intersecting this paraboloid with the coordinate planes at \(x = 0\), \(y = 0\), or \(z = 0\) gives us sections of the surface. For example, setting \(z = 0\) flattens the surface into a parabolic curve on the \(xy\)-plane.
- Equally, intersecting with \(x = 0\) or \(y = 0\) results in another set of simple shapes; a line and a parabola respectively.
Other exercises in this chapter
Problem 29
In Exercises \(21-30,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$ \int_{0}^{1} \int_{-
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The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders are possible and occasionally easier to eva
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Average height of a cone Find the average height of the (single) cone \(z=\sqrt{x^{2}+y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\) -plane.
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