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Question Number 211749 by universe last updated on 19/Sep/24

$$volumeboundedbythecurve$$$$z=\sqrt{3x^2+3y^2}and{x}^2+y^2+z^2=6^2$$

Answered by mr W last updated on 20/Sep/24

Commented by mr W last updated on 20/Sep/24

$$z=\sqrt{3(x^2+y^2)}=\sqrt{3}r$$$$h=\sqrt{3}r$$$${\left(\sqrt{3}r\right)}^2+r^2=R^2=6^2$$$$\Rightarrow r=3\Rightarrow h=3\sqrt{3}$$$$V=\frac{2\pi R^2(R-h)}{3}=\frac{2\pi \times 6^2\times 3(2-\sqrt{3})}{3}$$$$=72(2-\sqrt{3})\pi$$

Commented by universe last updated on 20/Sep/24

$$thankyousir$$

Commented by universe last updated on 20/Sep/24

$$canuexplainV=\frac{2\pi R^2(R-r)}{3}$$

Commented by mr W last updated on 20/Sep/24

Commented by universe last updated on 20/Sep/24

$$thankssir$$

Answered by BHOOPENDRA last updated on 20/Sep/24

Answered by BHOOPENDRA last updated on 20/Sep/24

$$V=\int \int \int {\rho }^{2}sin\varphi d\rho d\theta d\varphi$$$$={\int }_0^{\frac{\pi }{6}}{\int }_0^{2\pi }{\int }_0^6{\rho }^{2}sin\varphi d\rho d\theta d\varphi$$$$={\int }_0^{\frac{\pi }{6}}{\int }_0^{2\pi }\left(\frac{6^3-0^3}{3}\right)sin\varphi d\theta d\varphi$$$$={\int }_0^{\frac{\pi }{6}}72(2\pi -0)sin\varphi d\varphi$$$$=-144\pi cos\varphi {\mid }_0^{\frac{\pi }{6}}$$$$=144\pi \left(\frac{2-\sqrt{3}}{2}\right)$$$$=72\pi (2-\sqrt{3})$$

Commented by universe last updated on 20/Sep/24

$$thankyousir$$

Commented by mr W last updated on 20/Sep/24

$$V=72\pi (2-\sqrt{3})>0$$

Commented by BHOOPENDRA last updated on 20/Sep/24

$$ThanksMr.Witwastypo$$

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