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Question Number 83383 by mhmd last updated on 01/Mar/20

$$find{\int }_0^2{\int }_0^{\sqrt{4-x^2}}{\int }_0^{2-z}zdxdydz$$$$pleashelpmesir$$

Answered by mr W last updated on 01/Mar/20

$${\int }_0^2{\int }_0^{\sqrt{4-x^2}}{\int }_0^{2-z}zdxdydz$$$$={\int }_0^2{\int }_0^{\sqrt{4-x^2}}\left({\int }_0^{2-z}zdy\right)dzdx$$$$={\int }_0^2\left({\int }_0^{\sqrt{4-x^2}}z(2-z)dz\right)dx$$$$={\int }_0^2{[z^2-\frac{z^3}{3}]}_0^{\sqrt{4-x^2}}dx$$$$={\int }_0^2[4-x^2-\frac{{(4-x^2)}^{\frac{3}{2}}}{3}]dx$$$$=4\times 2-\frac{2^3}{3}-\frac{1}{3}{\int }_0^2{(4-x^2)}^{\frac{3}{2}}dx$$$$=\frac{16}{3}-\frac{1}{3}{\int }_0^2{(4-x^2)}^{\frac{3}{2}}dx$$$$=\frac{16}{3}-\frac{1}{3}\times 3\pi \left(seebelow\right)$$$$=\frac{16}{3}-\pi$$$$$$$$letx=2\sin \theta$$$${\int }_0^2{(4-x^2)}^{\frac{3}{2}}dx$$$$={\int }_0^{\frac{\pi }{2}}{\left(4{\cos }^2\theta \right)}^{\frac{3}{2}}2\cos \theta d\theta$$$$=16{\int }_0^{\frac{\pi }{2}}{\cos }^4\theta d\theta$$$$=4{\int }_0^{\frac{\pi }{2}}{\left(2{\cos }^2\theta \right)}^{2}d\theta$$$$=4{\int }_0^{\frac{\pi }{2}}{(1+\cos 2\theta )}^{2}d\theta$$$$=4{\int }_0^{\frac{\pi }{2}}(1+2\cos 2\theta +{\cos }^{2}2\theta )d\theta$$$$=4{\int }_0^{\frac{\pi }{2}}(1+2\cos 2\theta +\frac{1+\cos 4\theta }{2})d\theta$$$$=4{\int }_0^{\frac{\pi }{2}}(\frac{3}{2}+2\cos 2\theta +\frac{\cos 4\theta }{2})d\theta$$$$=4{[\frac{3}{2}\theta +\sin 2\theta +\frac{\sin 4\theta }{8}]}_0^{\frac{\pi }{2}}$$$$=4\times \frac{3}{2}\times \frac{\pi }{2}$$$$=3\pi$$

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