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Double Integral: int int (3x - 4x sqrt(xy)] dx, R={0<=x<=4,
- Thread starter Seimuna
- Start date
Hello, Seimuna!
\(\displaystyle \text{With respect to }x\!:\;\;\int^9_0\underbrace{\int^4_0\bigg(3x - 4y^{\frac{1}{2}}x^{\frac{3}{2}}\bigg)\,dx}\,dy\)
\(\displaystyle \text{Inside, we have: }\;\frac{3}{2}x^2 - \frac{8}{5}y^{\frac{1}{2}}x^{\frac{5}{2}}\,\bigg]^4_0 \;=\;\bigg(\frac{3}{2}\cdot16 - \frac{8}{5}y^{\frac{1}{2}}\cdot32\bigg) - \bigg(0 - 0\bigg) \;=\;24 - \frac{256}{5}y^{\frac{1}{2}}\)
\(\displaystyle \text{With respect to }y\!:\;\;\int^9_0\bigg(24 - \frac{256}{5}y^{\frac{1}{2}}\bigg)\,dy \;=\;24y - \frac{512}{15}y^{\frac{3}{2}}\,\bigg]^9_0\)
. . . . . . \(\displaystyle = \;\bigg(24\cdot9 - \frac{512}{15}\cdot27\bigg) - \bigg(0 - 0\bigg) \;=\;216 - \frac{13824}{15} \;=\;\boxed{-\frac{3528}{5}}\)
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\(\displaystyle \text{With respect to }y\!:\;\;\int^4_0\underbrace{\int^9_0\bigg(3x-4x^{\frac{3}{2}}y^{\frac{1}{2}}\bigg)\,dy}\,dx\)
\(\displaystyle \text{Inside, we have: }\;3xy - \frac{8}{3}x^{\frac{3}{2}}y^{\frac{3}{2}}\,\bigg]^9_0 \;=\;\bigg(3x\cdot9 - \frac{8}{3}x^{\frac{3}{2}}\cdot27\bigg) - \bigg(0 - 0\bigg) \;=\;27x - 72x^{\frac{3}{2}}\)
\(\displaystyle \text{With respect to }x\!:\;\;\int^4_0\left(27x - 72x^{\frac{3}{2}}\right)\,dx \;=\; \frac{27}{2}x^2 - \frac{144}{5}x^{\frac{5}{2}}\,\bigg]^4_0\)
. . . . . . \(\displaystyle = \;\bigg(\frac{27}{2}\cdot16 - \frac{144}{5}\cdot32\bihh) - \bigg(0 - 0\bigg) \;=\;216 -\frac{4608}{5} \;=\;\boxed{-\frac{3528}{5}}\)
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Hope you find your error(s).
There are certainly plenty of opportunities for them.
\(\displaystyle \int\int [3x-4x \sqrt{xy}]\,dA \quad\text{where }R \:=\:\{0 \leq x \leq 4,\;\;0 \leq y \leq 9\}\)
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.
\(\displaystyle \text{With respect to }x\!:\;\;\int^9_0\underbrace{\int^4_0\bigg(3x - 4y^{\frac{1}{2}}x^{\frac{3}{2}}\bigg)\,dx}\,dy\)
\(\displaystyle \text{Inside, we have: }\;\frac{3}{2}x^2 - \frac{8}{5}y^{\frac{1}{2}}x^{\frac{5}{2}}\,\bigg]^4_0 \;=\;\bigg(\frac{3}{2}\cdot16 - \frac{8}{5}y^{\frac{1}{2}}\cdot32\bigg) - \bigg(0 - 0\bigg) \;=\;24 - \frac{256}{5}y^{\frac{1}{2}}\)
\(\displaystyle \text{With respect to }y\!:\;\;\int^9_0\bigg(24 - \frac{256}{5}y^{\frac{1}{2}}\bigg)\,dy \;=\;24y - \frac{512}{15}y^{\frac{3}{2}}\,\bigg]^9_0\)
. . . . . . \(\displaystyle = \;\bigg(24\cdot9 - \frac{512}{15}\cdot27\bigg) - \bigg(0 - 0\bigg) \;=\;216 - \frac{13824}{15} \;=\;\boxed{-\frac{3528}{5}}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
\(\displaystyle \text{With respect to }y\!:\;\;\int^4_0\underbrace{\int^9_0\bigg(3x-4x^{\frac{3}{2}}y^{\frac{1}{2}}\bigg)\,dy}\,dx\)
\(\displaystyle \text{Inside, we have: }\;3xy - \frac{8}{3}x^{\frac{3}{2}}y^{\frac{3}{2}}\,\bigg]^9_0 \;=\;\bigg(3x\cdot9 - \frac{8}{3}x^{\frac{3}{2}}\cdot27\bigg) - \bigg(0 - 0\bigg) \;=\;27x - 72x^{\frac{3}{2}}\)
\(\displaystyle \text{With respect to }x\!:\;\;\int^4_0\left(27x - 72x^{\frac{3}{2}}\right)\,dx \;=\; \frac{27}{2}x^2 - \frac{144}{5}x^{\frac{5}{2}}\,\bigg]^4_0\)
. . . . . . \(\displaystyle = \;\bigg(\frac{27}{2}\cdot16 - \frac{144}{5}\cdot32\bihh) - \bigg(0 - 0\bigg) \;=\;216 -\frac{4608}{5} \;=\;\boxed{-\frac{3528}{5}}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Hope you find your error(s).
There are certainly plenty of opportunities for them.