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Question Number 215704 by universe last updated on 15/Jan/25

Commented by universe last updated on 16/Jan/25

$$HIMRWpleasesolvethisproblem$$

Answered by mr W last updated on 16/Jan/25

$$\int \int \int (x^2+y^2+z^2)dxdydz$$$$={\int }_0^a{\int }_0^{a-z}{\int }_0^{a-z-y}(x^2+y^2+z^2)dxdydz$$$$={\int }_0^a{\int }_0^{a-z}[\frac{{(a-z-y)}^3}{3}+y^2(a-z-y)+z^2(a-z-y)]dydz$$$$={\int }_0^a{\int }_0^{a-z}[\frac{{(a-z-y)}^3}{3}+(a-z)y^2-y^3+z^2(a-z)-z^{2}y]dydz$$$$={\int }_0^a[\frac{{(a-z)}^4}{3\times 4}+\frac{(a-z){(a-z)}^3}{3}-\frac{{(a-z)}^4}{4}+z^2(a-z)(a-z)-\frac{z^2{(a-z)}^2}{2}]dz$$$$={\int }_0^a[\frac{{(a-z)}^4}{6}+\frac{z^2{(a-z)}^2}{2}]dz$$$$={\int }_0^a[\frac{{(a-z)}^4}{6}+\frac{z^4-2{az}^3+a^2{z}^2}{2}]dz$$$$=\frac{1}{6}\times \frac{a^5}{5}+\frac{1}{2}(\frac{a^5}{5}-\frac{2a\times a^4}{4}+\frac{a^2\times a^3}{3})$$$$=\frac{a^5}{20}✓$$

Commented by universe last updated on 16/Jan/25

$$thankyouverymuchsir$$

Commented by universe last updated on 16/Jan/25

$$$$$$x=au,y=av,z=aw$$$$\int \int \int a^2(u^2+v^2+w^2)a^{3}dudvdw$$$$a^5\int \int \int u^{2}dudvdw+a^5\int \int \int v^{2}dudvdw+a^5\int \int \int w^{2}dudvdw$$$$3a^5\int \int \int u^{3-1}{v}^{1-1}{w}^{1-1}dudvdw$$$$byderichletintegral$$$$3a^5\frac{\lceil 3\lceil 1\lceil 1}{\lceil (3+1+1+1)}=3a^5\frac{2!}{5!}=a^5/20$$

Commented by universe last updated on 16/Jan/25

$$MRWsirwhatismistakeinthissolution??$$

Commented by mr W last updated on 16/Jan/25

$$thereisnomistakeinthissolution.$$$$imadeamistake.now{it}^{\prime }sfixed.$$$$ialsoget\frac{a^5}{20}.$$

Commented by universe last updated on 16/Jan/25

$$thankyousir$$

Commented by ajfour last updated on 17/Jan/25

https://youtu.be/KroW0v7N_Jg?si=8W3jW8TXO6USpeV1

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