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Chapter 17: Problem 7

Find the area enclosed by the graph of \(y=\frac{1}{\sqrt{9-4 t^{2}}}\) between \(t=0\) and \(t=1\).

Short Answer

Expert verified
Answer: The area enclosed by the graph of the function is \(A = \frac{1}{2}\arcsin\left(\frac{2}{3}\right)\).

Step by step solution

01

Set up the definite integral

To find the area enclosed by the given function between \(t=0\) and \(t=1\), we need to compute the following definite integral: $$ A = \int_{0}^{1}\frac{1}{\sqrt{9-4t^2}}dt $$
02

Apply substitution method

To solve the integral, we can use the substitution method. Let's use the substitution \(t=\frac{3}{2}\sin(u)\), then the differential \(dt = \frac{3}{2}\cos(u) du\). We also need to change the limits of integration accordingly: $$ t=0 \implies u=0 \\ t=1 \implies u=\arcsin\left(\frac{2}{3}\right) $$ Now, we can rewrite the integral as: $$ A = \int_{0}^{\arcsin\left(\frac{2}{3}\right)} \frac{1}{\sqrt{9-4\left(\frac{3}{2}\sin(u)\right)^2}}\cdot \frac{3}{2}\cos(u) du $$
03

Simplify the expression

We can simplify the expression inside the integral as follows: $$ A = \int_{0}^{\arcsin\left(\frac{2}{3}\right)} \frac{1}{\sqrt{9-9\sin^2(u)}}\cdot \frac{3}{2}\cos(u) du $$ The expression under the square root in the denominator becomes: $$ \sqrt{9-9\sin^2(u)} = \sqrt{9\cos^2(u)} = 3\cos(u) $$ Substituting and simplifying the expression: $$ A = \int_{0}^{\arcsin\left(\frac{2}{3}\right)} \frac{\frac{3}{2}\cos(u)}{3\cos(u)} du $$ Cancelling out the common terms, we get: $$ A = \int_{0}^{\arcsin\left(\frac{2}{3}\right)} \frac{1}{2} du $$
04

Evaluate the integral

Now, we can easily evaluate this simple integral as follows: $$ A = \frac{1}{2}\int_{0}^{\arcsin\left(\frac{2}{3}\right)} du = \frac{1}{2}[u]_{0}^{\arcsin\left(\frac{2}{3}\right)} = \frac{1}{2}(\arcsin\left(\frac{2}{3}\right)-0) $$ Thus, the area enclosed by the graph of the function is: $$ A = \frac{1}{2}\arcsin\left(\frac{2}{3}\right) $$

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