Answer:
The answer is "[tex]\frac{324}{5}[/tex]"
Step-by-step explanation:
If the missing value is this [tex]z= 9-y-x^2[/tex] then:
[tex]\bold{z =9-y-x^2} \\\\\to x\leq 0 \\\\\to y < 0\\\\let: \ z= 0\ then\\\\ \to 9-y-x^2=0\\\\\to y=9-x^2\\\\if\ y=0 \\\\\to 0=9-x^2\\\\\to x=3\\\\\text{Calculate the volume: }\\[/tex]
[tex]\to \int \int z dz\ \ = \int \int_{R} (9-y-x^2) dz\\\\[/tex]
[tex]=\int_0^{3} \ \int_0^{9-x^2} (9-y-x^2) dy \ dx\\\\=\int_0^{3} (9y- \frac{y^2}{2}-x^2 y)^{9-x^2} _{y=0}\ dx\\\\=\int_0^{3} (9-x^2)^2- \frac{(9-x^2)^2}{2}\ dx\\\\=\frac{1}{2} \int_0^{3} (9-x^2)^2\ dx\\\\=\frac{1}{2} \int_0^{3} 81+x^4-18x^2 \ dx\\\\=\frac{1}{2} (81x +\frac{x^5}{5}-6x^3)^3_0 \\\\= \frac{1}{2} (\frac{648}{5}) \\\\=\frac{324}{5}[/tex]
