Answer to Question #205524 in Calculus for Adnan Tariq

Question

Find volume bounded by the planes 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 and 2𝑥 + 𝑦 + 𝑧 = 6 0 ≤ 𝑥 ≤ 2,0 ≤ 𝑦 ≤ 6 − 2�

Answer 1

$$V=\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{6-2x}\displaystyle\int_{0}^{6-y-2x}dzdydx$$

$$=\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{6-2x}[z]\begin{matrix} 6-y-2x \\ 0 \end{matrix}dydx$$

$$=\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{6-2x}(6-y-2x)dydx$$

$$=\displaystyle\int_{0}^{2}[6y-\dfrac{y^2}{2}-2xy]\begin{matrix} 6-2x \\ 0 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{2}(36-12x-18+12x-2x^2-12x+4x^2)dx$$

$$=\displaystyle\int_{0}^{2}(2x^2-12x+18)dx$$

$$=[\dfrac{2}{3}x^3-6x^2+18x]\begin{matrix} 2 \\ 0 \end{matrix}$$

$$=\dfrac{16}{3}-24+36=\dfrac{52}{3}(units^3)$$

$V=\dfrac{52}{3}$ units of volume.