Double Integral: int int (3x - 4x sqrt(xy)] dx, R={0<=x<=4,

Seimuna

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[3x4xxy].dA\displaystyle \int\int [3x-4x \sqrt{xy}] .dA
where R = {0<=x<=4, 0<=y<=9}

for this, if i integrate with y first, the answer ll be different from if i integrate with x first... how shud i noe which 1 to integrate first ?
 
Hello, Seimuna!

[3x4xxy]dAwhere R={0x4,    0y9}\displaystyle \int\int [3x-4x \sqrt{xy}]\,dA \quad\text{where }R \:=\:\{0 \leq x \leq 4,\;\;0 \leq y \leq 9\}


if i integrate with y first, the answer will be different from if i integrate with x first.
This should not happen! . . . and you should know this.

With respect to x ⁣:    0904(3x4y12x32)dxdy\displaystyle \text{With respect to }x\!:\;\;\int^9_0\underbrace{\int^4_0\bigg(3x - 4y^{\frac{1}{2}}x^{\frac{3}{2}}\bigg)\,dx}\,dy

Inside, we have:   32x285y12x52]04  =  (321685y1232)(00)  =  242565y12\displaystyle \text{Inside, we have: }\;\frac{3}{2}x^2 - \frac{8}{5}y^{\frac{1}{2}}x^{\frac{5}{2}}\,\bigg]^4_0 \;=\;\bigg(\frac{3}{2}\cdot16 - \frac{8}{5}y^{\frac{1}{2}}\cdot32\bigg) - \bigg(0 - 0\bigg) \;=\;24 - \frac{256}{5}y^{\frac{1}{2}}


With respect to y ⁣:    09(242565y12)dy  =  24y51215y32]09\displaystyle \text{With respect to }y\!:\;\;\int^9_0\bigg(24 - \frac{256}{5}y^{\frac{1}{2}}\bigg)\,dy \;=\;24y - \frac{512}{15}y^{\frac{3}{2}}\,\bigg]^9_0

. . . . . . =  (2495121527)(00)  =  2161382415  =  35285\displaystyle = \;\bigg(24\cdot9 - \frac{512}{15}\cdot27\bigg) - \bigg(0 - 0\bigg) \;=\;216 - \frac{13824}{15} \;=\;\boxed{-\frac{3528}{5}}


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With respect to y ⁣:    0409(3x4x32y12)dydx\displaystyle \text{With respect to }y\!:\;\;\int^4_0\underbrace{\int^9_0\bigg(3x-4x^{\frac{3}{2}}y^{\frac{1}{2}}\bigg)\,dy}\,dx


Inside, we have:   3xy83x32y32]09  =  (3x983x3227)(00)  =  27x72x32\displaystyle \text{Inside, we have: }\;3xy - \frac{8}{3}x^{\frac{3}{2}}y^{\frac{3}{2}}\,\bigg]^9_0 \;=\;\bigg(3x\cdot9 - \frac{8}{3}x^{\frac{3}{2}}\cdot27\bigg) - \bigg(0 - 0\bigg) \;=\;27x - 72x^{\frac{3}{2}}


With respect to x ⁣:    04(27x72x32)dx  =  272x21445x52]04\displaystyle \text{With respect to }x\!:\;\;\int^4_0\left(27x - 72x^{\frac{3}{2}}\right)\,dx \;=\; \frac{27}{2}x^2 - \frac{144}{5}x^{\frac{5}{2}}\,\bigg]^4_0

. . . . . . \(\displaystyle = \;\bigg(\frac{27}{2}\cdot16 - \frac{144}{5}\cdot32\bihh) - \bigg(0 - 0\bigg) \;=\;216 -\frac{4608}{5} \;=\;\boxed{-\frac{3528}{5}}\)


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Hope you find your error(s).
There are certainly plenty of opportunities for them.

 
actually i do know about this... but for few question i got different answer... i wonder if im the 1 who make the mistake of taking it as the same answer...hence, i try to ask here... thanks...i ll check out wat i did wrongly...
 
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