Set up a triple integral for the volume of the solid bounded by the coordinate planes and the plane given below. $z = 12 - 6 x - 8 y$
A)$ \int_{0}^{2} \int_{0}^{\frac{12-6 x}{8}}\int_{0} ^ {12-6x-8y}dzdydx $

B) $ \int_{0}^{2} \int_{0}^{\frac{12-6 x}{8}}\int_{0} ^ {12-6x-8y}dydzdx $

C) $\int _ { 0 } ^ { 12 } \int _ { 0 } ^ { \frac { 12 - 6 x } { 8 } } \int _ { 0 } ^ { 12 - 6 x - 8 y } d z d y d x$

D) $\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { \frac { 12 - 6 x } { 8 } } \int _ { 0 } ^ { 12 - 6 x - 8 y } d x d y d z$

E)$ \int_{0}^{\frac{12-6x}{8}} \int_{0}^{2}\int_{0} ^ {12-6x-8y}dzdydx $

Question

Set up a triple integral for the volume of the solid bounded by the coordinate planes and the plane given below. $z = 12 - 6 x - 8 y$
A)$ \int_{0}^{2} \int_{0}^{\frac{12-6 x}{8}}\int_{0} ^ {12-6x-8y}dzdydx $

B) $ \int_{0}^{2} \int_{0}^{\frac{12-6 x}{8}}\int_{0} ^ {12-6x-8y}dydzdx $

C) $\int _ { 0 } ^ { 12 } \int _ { 0 } ^ { \frac { 12 - 6 x } { 8 } } \int _ { 0 } ^ { 12 - 6 x - 8 y } d z d y d x$

D) $\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { \frac { 12 - 6 x } { 8 } } \int _ { 0 } ^ { 12 - 6 x - 8 y } d x d y d z$

E)$ \int_{0}^{\frac{12-6x}{8}} \int_{0}^{2}\int_{0} ^ {12-6x-8y}dzdydx $

Quizplus · Accepted Answer

The solid is bounded by the coordinate planes (x=0, y=0, z=0) and the plane $z = 12 - 6x - 8y$. To find the volume, we integrate in the z-direction first, from 0 to $12 - 6x - 8y$, then in the y-direction, and finally in the x-direction. The limits for x are from 0 to 2, as found by setting z=0 in the plane equation and solving for x when y=0, giving $x=2$. The limits for y are from 0 to $\frac{12-6x}{8}$, found by setting z=0 and solving for y in terms of x. This setup correctly accounts for the volume under the plane and above the xy-plane within the first octant, hence choice A is correct.

Set Up a Triple Integral for the Volume of the Solid