Use a triple integral to find the volume of the solid bounded by $x = y ^ { 2 }$ and the planes $z = 0$ and $x + z = 3$.
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$$\int_0^9 \int_0^{\sqrt{x}} \int_0^{3-x} dz \ dy \ dx = \int_0^9 \int_0^{\sqrt{x}} (3-x) \ dy \ dx = \int_0^9 \left[ 3y - xy \right]_0^{\sqrt{x}} \ dx = \int_0^9 (3\sqrt{x} - x\sqrt{x}) \ dx $$$$= \left[ 2x^{3/2} - \frac{2}{5}x^{5/2} \right]_0^9 = 2(3^3) - \frac{2}{5}(3^5) = 54 - \frac{486}{5} = \frac{270}{5} = 54$$

Use a Triple Integral to Find the Volume of the Solid