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Question: Find the volume bounded by the cylinder $x^2+y^2=4$ and the planes $y+z=a$ and $z=0\cdot(H)$
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Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : High

am2(82) • 223 views
 modified 5 months ago by Juilee • 1.8k written 7 months ago by
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z varies from z = 0 to $z = a-y = a – r sin \theta$

$\theta$ varies from 0 to 2$\pi$ and r varies from 0 to 2

$V = \int^2_0 \int^{2\pi}_0 \int^{arsin\theta}_0 dz \hspace{0.1cm}r\hspace{0.1cm}dr\hspace{0.1cm}d \theta\\ \hspace{0.2cm} = \int^{2a}_{r=0} \int^{2\pi}_{\theta = 0} [z]^{a-rsin\theta}_0 dr \hspace{0.1cm}d\theta = \int^{2a}_0 \int^{2\pi}_0(a-rsin\theta) r \hspace{0.1cm} dr\hspace{0.1cm} d\theta\\ \hspace{0.2cm} = \int^{2a}_0(a\theta+ rcos\theta)^{2\pi}_0r \hspace{0.1cm}dr=\int_0^{2a}(2 \pi a+r(1-1))r \hspace{0.1cm}dr\\ \hspace{0.2cm} = 2 \pi a \int_0^2r \hspace{0.1cm}dr = 2 \pi a\big[\frac{r^2}{2}\big]_0^2=4 \pi a$