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Magazine Chapter 14: Problem 10
Find the surface area of the given surface. The portion of the graph of \(z=\frac{2}{3} \sqrt{2} x^{3 / 2}+\frac{2}{3} y^{3 / 2}\) directly over the region in the \(x y\) plane between the graph of \(y=x^{2}\) and the \(x\) axis on \([0,1]\)
Short Answer
Expert verified
The surface area requires the evaluation of a non-trivial integral, which may need numerical methods for exact evaluation.
Step by step solution
01
Understand the problem
We need to find the surface area of a part of the surface defined by the function \(z = \frac{2}{3} \sqrt{2} x^{3/2} + \frac{2}{3} y^{3/2}\). The region of interest is over the area in the \(xy\)-plane between \(y = x^2\) and the \(x\)-axis, from \(x = 0\) to \(x = 1\). This means the region can be described as \(0 \leq y \leq x^2\) and \(0 \leq x \leq 1\).
02
Set up the surface area integral
The surface area of a surface given by \(z = f(x, y)\) over a region \(R\) in the \(xy\)-plane is given by the double integral \[ \int \int_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dy \, dx. \] First, we must find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) for our function \(z\).
03
Calculate partial derivatives
Differentiate \(z\) with respect to \(x\):\[ \frac{\partial z}{\partial x} = \frac{2}{3} \cdot \frac{3}{2} \sqrt{2} x^{1/2} = \sqrt{2} x^{1/2}. \]Now differentiate \(z\) with respect to \(y\):\[ \frac{\partial z}{\partial y} = \frac{2}{3} \cdot \frac{3}{2} y^{1/2} = y^{1/2}. \]
04
Plug into surface area formula
Substitute the partial derivatives into the surface area formula:\[ \sqrt{1 + \left( \sqrt{2} x^{1/2} \right)^2 + \left( y^{1/2} \right)^2} = \sqrt{1 + 2x + y}. \]Thus, the surface area integral becomes:\[ \int_0^1 \int_0^{x^2} \sqrt{1 + 2x + y} \, dy \, dx. \]
05
Compute the inner integral
Compute the inner integral with respect to \(y\):\[ \int_0^{x^2} \sqrt{1 + 2x + y} \, dy. \]Let \(u = 1 + 2x + y\), then \(du = dy\), and when \(y=0\), \(u = 1 + 2x\), when \(y=x^2\), \(u = 1 + 2x + x^2\).Thus, the integral becomes:\[ \int_{1 + 2x}^{1 + 2x + x^2} \sqrt{u} \, du = \left[ \frac{2}{3} u^{3/2} \right]_{1 + 2x}^{1 + 2x + x^2}. \]
06
Evaluate the antiderivative
Evaluate the expression:\[ \left[ \frac{2}{3} (1 + 2x + x^2)^{3/2} \right] - \left[ \frac{2}{3} (1 + 2x)^{3/2} \right]. \]This yields \(\frac{2}{3} [(1 + 2x + x^2)^{3/2} - (1 + 2x)^{3/2}]\).
07
Compute the outer integral
Substitute the result of the inner integral into the outer integral:\[ \int_0^1 \frac{2}{3} [(1 + 2x + x^2)^{3/2} - (1 + 2x)^{3/2}] \, dx. \]This integral can be evaluated using numerical methods or by using a calculator since it may not have a simple antiderivative.
08
Final surface area
The evaluated integral gives the surface area. If solved numerically, the solution can be approximated to a specific decimal point, depending on the required accuracy of the answer.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives help us find the rate of change of a function with respect to one variable, while keeping the others constant. In the given exercise, we have a surface described by the function: \[ z = \frac{2}{3} \sqrt{2} x^{3/2} + \frac{2}{3} y^{3/2}. \]
- To find the partial derivative with respect to \(x\), we treat \(y\) as a constant. The derivative is: \[ \frac{\partial z}{\partial x} = \sqrt{2} x^{1/2}. \]
- For the partial derivative with respect to \(y\), \(x\) is treated as a constant, yielding:\[ \frac{\partial z}{\partial y} = y^{1/2}. \]
Surface Area Integral
The surface area of a function over a region can be calculated using a surface area integral. For a function\( z = f(x, y) \), the surface area over a region \( R \) in the \( xy \)-plane is:\[\int \int_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dy \, dx.\]Here, we substititued the partial derivatives into the formula making it:\[ \sqrt{1 + 2x + y} \]resulting in the integral:\[ \int_0^1 \int_0^{x^2} \sqrt{1 + 2x + y} \, dy \, dx.\]This setup is crucial for determining the surface area in the desired region of the \( xy \)-plane.
Calculus Integration
Calculus integration involves finding the integral of a function, which can signify the area under a curve or, in our case, the surface area. For the inner integral in the exercise, we use a substitution:
- Let \( u = 1 + 2x + y \), thus \( du = dy \).
- This transforms the limits from \( y = 0 \) to \( u = 1 + 2x \), and \( y = x^2 \) to \( u = 1 + 2x + x^2 \).
Numerical Methods
Numerical methods are invaluable for integrals without simple antiderivatives. In the exercise, the outer integral:\[ \int_0^1 \frac{2}{3} [(1 + 2x + x^2)^{3/2} - (1 + 2x)^{3/2}] \, dx \]does not have an obvious solution through elementary functions.This is where numerical integration, like the Trapezoidal Rule or Simpson鈥檚 Rule, comes into play.
- These methods approximate the integral by breaking it into small intervals, calculating area contributions and summing them.
- Modern calculators or software (like Python or MATLAB) can also compute such integrals accurately.
