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Question: Find the volume in the first octant bounded by the cylinder $x^2 + y^2 = 2$ and the planes $z = x + y,\, y = x,\, z = 0$ and $x = 0(H)$

Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : High

am2(82) • 77 views
modified 5 days ago by gravatar for Juilee Juilee1.6k written 10 weeks ago by gravatar for smitapn612 smitapn6120

If we take projections on the XY plane, the area is bounded by the circle $x^2 + y^2 = 2$, the line y = x and the line x = 0.

We change the coordinate to cylindrical polar by putting x = $rcos \theta, y = rsin \theta$, z = z .

Eqn. of the cylinder becomes $x^2 + y^2 = 2 \Rightarrow r = \sqrt{2}$ then the line becomes $y= x \Rightarrow \theta = \pi/4$

$x=0 $ becomes $\theta = \frac{\pi}{2}$

If we consider a radial strip in the projection, r varies from $0 \hspace{0.1cm}\text{to} \sqrt{2}$, $\theta$ varies from $\theta = \frac{\pi}{4} \hspace{0.1cm}\text{to}\hspace{0.1cm}\theta = \frac{\pi}{2}$ then z varies from z = 0 to z = x + y

$V = \int_{\theta = \pi / 4}^{\pi/2}\int_{r = 0}^{\sqrt{2}}\int_{z=0}^{r(cos \theta+sin \theta)} r \hspace{0.1cm}dr\hspace{0.1cm}d\theta\hspace{0.1cm}dz\\ \hspace{0.2cm}= \int_{\theta = \pi / 4}^{\pi/2}\int_{r = 0}^{\sqrt{2}}r [z]_0^{r(cos \theta+sin \theta)}dr\hspace{0.1cm}d\theta\\ \hspace{0.2cm}= \int_{\theta = \pi / 4}^{\pi/2}\int_{r = 0}^{\sqrt{2}}r^2(cos \theta + sin \theta)dr\hspace{0.1cm}d\theta\\ \hspace{0.2cm}= \int_{\theta= \pi/4}^{\pi/2}(cos \theta + sin \theta)\Big[\frac{r^3}{3}\Big]_0^{\sqrt{2}}d \theta\\ \hspace{0.1cm} = \frac{2 \sqrt{2}}{3}\int_{\theta=\pi/4}^{\pi/2}(cos \theta + sin \theta)d \theta = \frac{2 \sqrt{2}}{3}\Big[sin \theta - cos \theta\Big]_{\pi/4}^{\pi/2}\\ \hspace{0.2cm} = \frac{2 \sqrt{2}}{3}\Big[(1-0) - \big(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\big)\Big] = \frac{2 \sqrt{2}}{3}$

written 5 days ago by gravatar for Juilee Juilee1.6k
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