Solve the problem.
-Write an iterated triple integral in the order $\mathrm { dz } \mathrm { dy } \mathrm {} \mathrm { dx }$ for the volume of the tetrahedron cut from the first octant by the plane $\frac { x } { 10 } + \frac { y } { 3 } + \frac { z } { 5 } = 1$.
A) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 1 - y / 3 } \int _ { 0 } ^ { 1 - x / 10 - y / 3 } d z d y d x$
B) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 1 - x / 10 } \int _ { 0 } ^ { 1 - x / 10 - y / 3 } d z d y d x$
C)$\int_{0}^{10} \int_{0}^{3(1-x / 10)} \int_{0}^{5(1-x / 10-y / 3)} d z d y d x$
D)$\int_{0}^{10} \int_{0}^{10(1-y / 3)} \int_{0}^{5(1-x / 10-y / 3)} d z d y d x$

Question

Solve the problem.
-Write an iterated triple integral in the order $\mathrm { dz } \mathrm { dy } \mathrm {} \mathrm { dx }$ for the volume of the tetrahedron cut from the first octant by the plane $\frac { x } { 10 } + \frac { y } { 3 } + \frac { z } { 5 } = 1$.
A) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 1 - y / 3 } \int _ { 0 } ^ { 1 - x / 10 - y / 3 } d z d y d x$
B) $\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { 1 - x / 10 } \int _ { 0 } ^ { 1 - x / 10 - y / 3 } d z d y d x$
C)$\int_{0}^{10} \int_{0}^{3(1-x / 10)} \int_{0}^{5(1-x / 10-y / 3)} d z d y d x$ 
D)$\int_{0}^{10} \int_{0}^{10(1-y / 3)} \int_{0}^{5(1-x / 10-y / 3)} d z d y d x$

Quizplus · Accepted Answer

The equation of the plane can be rearranged as: 
$$z = 5 - \frac{1}{10}x - \frac{1}{3}y$$
The tetrahedron is cut from the first octant, which means it is bounded by the coordinate planes x = 0, y = 0, and z = 0, as well as by the plane above.

Therefore, the limits of integration for x, y, and z are:

$$0 \leq x \leq 10$$
$$0 \leq y \leq 3\left(1-\frac{x}{10}\right)$$
$$0 \leq z \leq 5\left(1-\frac{x}{10}-\frac{y}{3}\right)$$

So the iterated triple integral in the order dz dy dx is:

$$\int_0^{10} \int_0^{3\left(1-\frac{x}{10}\right)} \int_0^{5\left(1-\frac{x}{10}-\frac{y}{3}\right)} dzdydx = \boxed{\textbf{(C)}}$$

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