Use a triple integral to find the volume of the solid bounded by $x = y ^ { 2 }$ and the planes $z = 0$ and $x + z = 3$.
Select the correct answer.
A) $8.3$
B) $18.3$
C) $2.5$
D) $15.3$
E) $11.3$

Question

Quizplus · Accepted Answer

The limits of integration are $0 \le z \le 3 - x$, $0 \le x \le 9$, and $0 \le y \le \sqrt{x}$. So, the volume is$$\int_{0}^{9} \int_{0}^{\sqrt{x}} \int_{0}^{3-x} dz dy dx = \int_{0}^{9} \int_{0}^{\sqrt{x}} (3-x) dy dx = \int_{0}^{9} \left[ 3y - xy \right]_{0}^{\sqrt{x}} dx = \int_{0}^{9} (3\sqrt{x} - x\sqrt{x}) dx = \left[ 2x^{3/2} - \frac{2}{5}x^{5/2} \right]_{0}^{9} = 18.3$$

Use a Triple Integral to Find the Volume of the Solid