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Chapter 17: Problem 32

Surface area using an explicit description Find the area of the following surfaces using an explicit description of the surface. $$\text { The trough } z=\frac{1}{2} x^{2}, \text { for }-1 \leq x \leq 1,0 \leq y \leq 4$$

Short Answer

Expert verified
Create a short answer question: Question: Find the surface area of the trough defined by the function \(z = \frac{1}{2} x^2\) over the region \(-1 \leq x \leq 1\) and \(0 \leq y \leq 4\). Answer: The surface area of the trough is \(8\sqrt{2}\).

Step by step solution

01

To find the surface area using an explicit description, we first need to find the partial derivatives of the function \(z\) with respect to \(x\) and \(y\). In this case, we have \(z = \frac{1}{2} x^2\). So, the partial derivatives are: $$ \frac{\partial z}{\partial x} = x, \qquad \frac{\partial z}{\partial y} = 0 $$ #Step 2: Calculate the expression for the surface area

Next, we will use the formula for the surface area of an explicitly defined surface: $$ dS = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dxdy $$ Plugging in the partial derivatives we calculated in the previous step, we get: $$ dS = \sqrt{1 + x^2 + 0^2} dx dy $$ #Step 3: Set up the integrals for the surface area
02

Now, we need to set up the integrals for the surface area using the expression we calculated for \(dS\). Since our region is defined by \(-1 \leq x \leq 1\) and \(0 \leq y \leq 4\), our integrals will be as follows: $$ S = \int_{0}^{4} \int_{-1}^{1} \sqrt{1 + x^2} dx dy $$ #Step 4: Integrate and calculate the surface area

Finally, we will evaluate the double integral to find the surface area. We first integrate with respect to \(x\): $$ S = \int_{0}^{4} \left[ \frac{x\sqrt{1 + x^2}}{2} + \frac{1}{2}\sinh^{-1}(x) \right]_{-1}^{1} dy $$ Evaluating the expression at the limits, we get: $$ S = \int_{0}^{4} \left(\frac{2\sqrt{2}}{2} - \frac{1}{2}\sinh^{-1}(1) - \frac{-2\sqrt{2}}{2} + \frac{1}{2}\sinh^{-1}(1) \right) dy $$ Simplifying, we have: $$ S = \int_{0}^{4} (2\sqrt{2}) dy $$ Now, we integrate with respect to \(y\): $$ S = \left[2\sqrt{2}y\right]_{0}^{4} $$ Evaluating the expression at the limits, we get the surface area of the trough: $$ S = 8\sqrt{2} $$

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Key Concepts

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

Partial Derivatives
In the context of surface area calculation, partial derivatives help us understand how our surface, described by a function, changes. Given a function like \( z = \frac{1}{2}x^2 \), taking the partial derivative with respect to \( x \) and \( y \) helps reveal these changes.

For \( z = \frac{1}{2}x^2 \):
  • \( \frac{\partial z}{\partial x} = x \) shows the rate of change of \( z \) as \( x \) changes, assuming \( y \) is constant.
  • \( \frac{\partial z}{\partial y} = 0 \) means \( z \) doesn't change as \( y \) varies since \( y \) isn鈥檛 in the expression.
This understanding simplifies our surface area formula and makes integration possible.
Double Integral
To find the surface area, we use a double integral, which essentially "adds up" tiny elements over a region. The adaptation of this, after considering partial derivatives, leads us to an expression:
\[ dS = \sqrt{1 + x^2 + 0^2} \, dx \, dy \]
This expression gets integrated over the given range for \( x \) and \( y \).
  • First, integrate with respect to \( x \).
  • Then, integrate the result with respect to \( y \).
This layered integration approach allows us to tackle complex shapes and calculate areas accurately. Each integral contribution accumulates to form the entire surface area.
Explicit Surface Description
An explicit surface description, like \( z = \frac{1}{2}x^2 \), provides a clear rule for calculating \( z \) based on \( x \) and \( y \).
  • This makes it straightforward to differentiate and integrate.
  • It removes ambiguity by defining a precise mathematical relationship.
Such descriptions are ideal for computations as they allow for direct manipulation and help form equations necessary for further analysis.
Having an explicit surface description simplifies the steps in solving for surface area and makes these tasks tractable with calculus tools.
Integration Limits
Integration limits specify the region over which you calculate the surface area. For our exercise, these gave us
  • \(-1 \leq x \leq 1\) and
  • \(0 \leq y \leq 4\)
as the boundaries of our integration process.

Setting Up Limits:

These limits define the range of \( x \) and \( y \) to consider.
  • They ensure that we're calculating the area precisely within the specified region.
  • Help maintain accuracy and relevance to the physical problem.
Properly understanding and setting these limits is crucial in achieving correct integration results that represent the physical surface identified by the given criteria.

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Most popular questions from this chapter

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) \(T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; D\) is the sphere of radius \(a\) centered at the origin.

Find the general formula for the surface area of a cone with height \(h\) and base radius \(a\) (excluding the base).

Cone and sphere The cone \(z^{2}=x^{2}+y^{2},\) for \(z \geq 0,\) cuts the sphere \(x^{2}+y^{2}+z^{2}=16\) along a curve \(C\) a. Find the surface area of the sphere below \(C,\) for \(z \geq 0\) b. Find the surface area of the sphere above \(C\) c. Find the surface area of the cone below \(C\), for \(z \geq 0\)

What's wrong? Consider the radial field \(\mathbf{F}=\frac{\langle x, y\rangle}{x^{2}+y^{2}}\) a. Verify that the divergence of \(\mathbf{F}\) is zero, which suggests that the double integral in the flux form of Green's Theorem is zero. b. Use a line integral to verify that the outward flux across the unit circle of the vector field is \(2 \pi\) c. Explain why the results of parts (a) and (b) do not agree.

Heat flux The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k > 0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\) \(T(x, y, z)=100 e^{-x-y} ; S\) consists of the faces of the cube \(|x| \leq 1\) \(|y| \leq 1,|z| \leq 1\)
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