Question: The density at any point of a cardioide $r=a (1 + cos\theta)$ varies as the square of its distance from its axis of symmetry. Find its mass.
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Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : Low

am2(82) • 326 views
Let $p(r, \theta)$ be any point on the given cardiode. The distance of p from it axis is $y = rsin \theta$. The density at any point is $p = kr^2sin^2 \theta$.
Mass of the lamina $= \int^{\pi}_{\theta = 0}\int_0^{a(1+cos \theta)}Kr^2 sin^2 \theta r \hspace{0.1cm}dr \hspace{0.1cm}d \theta\\ = 2k \int_0^\pi sin^2 \theta\Big[\frac{r^4}{4}\Big]_0^{a(1+cos \theta)}d \theta\\ = \frac{2k}{4} \int_0^\pi sin^2 \theta [a^4(1+cos \theta)^4]d \theta\\ = \frac{Ka^4}{2}\int_0^\pi sin^2 \theta (1 + cos \theta)^4 d \theta\\ = \frac{Ka^4}{2}\int_0^\pi(2 sin \theta/2 \hspace{0.1cm}cos \theta/2)^2(2 cos^2 \theta/2)^4 d \theta\\ = 32Ka^4\int^\pi_0sin^2 \theta/2 \hspace{0.1cm} cos^{10}\theta /2 \hspace{0.2cm} d \theta$ $\hspace{0.3cm}\text{Put} \hspace{0.2cm} \theta/2 = t \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \theta = 2t \hspace{0.3cm} \Rightarrow \hspace{0.3cm} d\theta = 2dt\\ \hspace{2.8cm}= 32Ka^4\int^{\pi/2}_0 sin^2 t cos^{10}t \hspace{0.1cm}2dt\\ \hspace{2.8cm}= 64Ka^4\int^{\pi/2}_0 sin^2 t cos^{10}t \hspace{0.1cm}dt\\ \hspace{2.8cm}= 32Ka^4 \beta\Big(\frac{3}{2}-\frac{11}{2}\Big)\\ \hspace{2.8cm}= 32Ka^4 \frac{\sqrt{\frac{3}{2}}\sqrt{\frac{11}{2}}}{\sqrt{7}}\\ \hspace{2.8cm}= 32Ka^4 \frac{\frac{1}{2}\sqrt{1/2}\hspace{0.1cm}\times \hspace{0.1cm}9/2 \hspace{0.1cm}\times \hspace{0.1cm}7/2 \hspace{0.1cm} \times \hspace{0.1cm} 5/2\hspace{0.1cm}\times \hspace{0.1cm} 3/2 \hspace{0.1cm}\times \hspace{0.1cm}1/2 \sqrt{1/2}}{6!}\\ \hspace{2.8cm}= 32Ka^4 \frac{9 \times 7 \times 5 \times 3 \times 1 \hspace{0.3cm}\pi}{2^6 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\\ \hspace{2.8cm} = 32 Ka^4 \frac{21 \hspace{0.3cm}\pi}{1024} = \frac{21}{32}a^4 \pi K$