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Chapter 6: Problem 57

Find the centroid of the region determined by the graphs of the inequalities. $$ y \leq 3 / \sqrt{x^{2}+9}, y \geq 0, x \geq-4, x \leq 4 $$

Short Answer

Expert verified
The centroid (ave(x), ave(y)) can be found by evaluating the integrals in steps 2, 3 and 4.

Step by step solution

01

Set Up the Region

The region R can be illustrated on an x-y graph as the area bounded by the curve y = 3 / 鈭�(x虏 + 9), the x-axis (y = 0) and the vertical lines x = -4, x = 4.
02

Calculate the Area

To find the area (A) of the region R, we can use the formula \(A = \int_{a}^{b} f(x)dx\). Here, \(f(x) = 3 / \sqrt{x^{2}+9}\) and the integration limit a = -4 and b = 4. So, the area is \(A = \int_{-4}^{4} 3 / \sqrt{x^{2}+9} dx\).
03

Calculate the x-coordinate

The formula for ave(x) is \(1/A \int_{a}^{b} xdA\). Substituting our limits a (-4) and b(4), and our function for dA (3/\sqrt{x^{2}+9}), we get: \(ave(x) = \frac{1}{A} \int_{-4}^{4} x * (3 / \sqrt{x^{2}+9}) dx\).
04

Calculate the y-coordinate

The formula for ave(y) is \(1/2A \int_{a}^{b} (f(x))^2 dx\). Substitute our limits a (-4) and b(4), and our function for dA (3/\sqrt{x^{2}+9}), we get: \(ave(y) = \frac{1}{2A} \int_{-4}^{4} (3 / \sqrt{x^{2}+9})^2 dx\).
05

Evaluate the Integral

After setting up the integrals in step 2, 3 and 4, evaluate the integrals to get the area (A) and the x and y coordinates.

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Most popular questions from this chapter

Graphical Analysis In Exercises \(\mathbf{6 1}\) and 62, graph \(f(x) / g(x)\) and \(f^{\prime}(x) / g^{\prime}(x)\) near \(x=0 .\) What do you notice about these ratios as \(x \rightarrow 0\) ? How does this illustrate L'H么pital's Rule? \(f(x)=\sin 3 x, \quad g(x)=\sin 4 x\)

Prove that if \(f(x) \geq 0, \lim _{x \rightarrow a} f(x)=0,\) and \(\lim _{x \rightarrow a} g(x)=\infty,\) then \(\lim _{x \rightarrow a} f(x)^{g(x)}=0\).

(a) Let \(f^{\prime}(x)\) be continuous. Show that \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h}=f^{\prime}(x)\) (b) Explain the result of part (a) graphically.

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Use integration by parts to verify the reduction formula. $$ \int \sec ^{n} x d x=\frac{1}{n-1} \sec ^{n-2} x \tan x+\frac{n-2}{n-1} \int \sec ^{n-2} x d x $$
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