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Misalkan $S$ merupakan permukaan \$$z=\\var{a}-3y+x^2\$$ yang berada di atas segitiga dengan titik-titik sudut \$$(0,0)\$$, \$$(2,0)\$$, \$$(2,-4)\$$.

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Jika \$$\\displaystyle \\iint_S z+3y-x^2\\,dS=\\frac{\\var{a}}{6}(a\\sqrt{a}-b\\sqrt{b})\$$, maka \$$a-b=\\ldots\$$.

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