Question

1.Use a triple integral to find the volume of the region bounded by z= 7x+8y, z=9, and the planes x=0 and y=0. Give the exact answer in the form of a fraction.

Answer 1

$V=\iiint\limits_V dv=\int_0^{\frac{9}{7}} dx\int\limits^{\frac{9}{8}-\frac{7}{8}x}_0 dy \int\limits_{7x+8y}^9 dz=$

$=\int_0^{\frac{9}{7}} dx\int\limits^{\frac{9}{8}-\frac{7}{8}x}_0 (9-(7x+8y))dy=\int_0^{\frac{9}{7}} \left((9-7x)(\frac{9}{8}-\frac{7}{8}x)-4(\frac{9}{8}-\frac{7}{8}x)^2\right)dx=$

=$\left( \frac{81 x}{16} - \frac{63 x^2}{16} +\frac{49 x^3}{48}\right)_0^{\frac{9}{7}}=\frac{243}{112}$