Question

Find the volume of the following solid. The solid in the first octant bounded by the coordinate planes and the surface zy

Answer 1

Answer:

The answer is "$\frac{324}{5}$"

Step-by-step explanation:

If the missing value is this $z= 9-y-x^2$ then:

$\bold{z =9-y-x^2} \\\\\to x\leq 0 \\\\\to y < 0\\\\let: \ z= 0\ then\\\\ \to 9-y-x^2=0\\\\\to y=9-x^2\\\\if\ y=0 \\\\\to 0=9-x^2\\\\\to x=3\\\\\text{Calculate the volume: }$

$\to \int \int z dz\ \ = \int \int_{R} (9-y-x^2) dz$

                  $=\int_0^{3} \ \int_0^{9-x^2} (9-y-x^2) dy \ dx\\\\=\int_0^{3} (9y- \frac{y^2}{2}-x^2 y)^{9-x^2} _{y=0}\ dx\\\\=\int_0^{3} (9-x^2)^2- \frac{(9-x^2)^2}{2}\ dx\\\\=\frac{1}{2} \int_0^{3} (9-x^2)^2\ dx\\\\=\frac{1}{2} \int_0^{3} 81+x^4-18x^2 \ dx\\\\=\frac{1}{2} (81x +\frac{x^5}{5}-6x^3)^3_0 \\\\= \frac{1}{2} (\frac{648}{5}) \\\\=\frac{324}{5}$