QuestionJuly 16, 2025

Find the area of the region that lies inside both curves. r=3sin(Theta ), r=3cos(Theta )

Find the area of the region that lies inside both curves. r=3sin(Theta ), r=3cos(Theta )
Find the area of the region that lies inside both curves. r=3sin(Theta ), r=3cos(Theta )

Solution
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Answer

\frac{9\pi}{8} Explanation 1. Identify Intersection Points Set 3\sin(\Theta) = 3\cos(\Theta) to find intersections. Simplify to \tan(\Theta) = 1, giving \Theta = \frac{\pi}{4} and \Theta = \frac{5\pi}{4}. 2. Calculate Area of Each Curve The area inside one curve is \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\Theta. For r=3\sin(\Theta), integrate from 0 to \frac{\pi}{4}. For r=3\cos(\Theta), integrate from 0 to \frac{\pi}{4}. 3. Integrate for r=3\sin(\Theta) \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (3\sin(\Theta))^2 \, d\Theta = \frac{9}{2} \int_{0}^{\frac{\pi}{4}} \sin^2(\Theta) \, d\Theta. Use \sin^2(\Theta) = \frac{1 - \cos(2\Theta)}{2}. 4. Integrate for r=3\cos(\Theta) \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (3\cos(\Theta))^2 \, d\Theta = \frac{9}{2} \int_{0}^{\frac{\pi}{4}} \cos^2(\Theta) \, d\Theta. Use \cos^2(\Theta) = \frac{1 + \cos(2\Theta)}{2}. 5. Compute Total Area Add the areas from both integrals calculated in Steps 3 and 4.

Explanation

1. Identify Intersection Points<br /> Set $3\sin(\Theta) = 3\cos(\Theta)$ to find intersections. Simplify to $\tan(\Theta) = 1$, giving $\Theta = \frac{\pi}{4}$ and $\Theta = \frac{5\pi}{4}$.<br /><br />2. Calculate Area of Each Curve<br /> The area inside one curve is $\frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\Theta$. For $r=3\sin(\Theta)$, integrate from $0$ to $\frac{\pi}{4}$. For $r=3\cos(\Theta)$, integrate from $0$ to $\frac{\pi}{4}$.<br /><br />3. Integrate for $r=3\sin(\Theta)$<br /> $\frac{1}{2} \int_{0}^{\frac{\pi}{4}} (3\sin(\Theta))^2 \, d\Theta = \frac{9}{2} \int_{0}^{\frac{\pi}{4}} \sin^2(\Theta) \, d\Theta$. Use $\sin^2(\Theta) = \frac{1 - \cos(2\Theta)}{2}$.<br /><br />4. Integrate for $r=3\cos(\Theta)$<br /> $\frac{1}{2} \int_{0}^{\frac{\pi}{4}} (3\cos(\Theta))^2 \, d\Theta = \frac{9}{2} \int_{0}^{\frac{\pi}{4}} \cos^2(\Theta) \, d\Theta$. Use $\cos^2(\Theta) = \frac{1 + \cos(2\Theta)}{2}$.<br /><br />5. Compute Total Area<br /> Add the areas from both integrals calculated in Steps 3 and 4.
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