Question

Let 𝑭 = 𝒙𝒛𝒊 − 𝒙𝒋 + 𝒚 𝟐𝒌. Evaluate ∭ 𝑭𝒅𝑽 𝑽 where V is the region bounded by the surfaces 𝒙 = 𝟎, 𝒚 = 𝟎, 𝒚 = 𝟏, 𝒛 = 𝒙 𝟐 , 𝒛 = 𝟏

Answer 1

$$\int\int\int_V\vec Fdv=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\displaystyle\int_{x^2}^{1}(xz\vec i-x\vec j+y^2\vec k)dzdydx$$

$$=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\big[\dfrac{xz^2}{2}\vec i-xz\vec j+y^2z\vec k\big]\begin{matrix} 1 \\ x^2 \end{matrix}dydx$$

$$=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\big[\dfrac{1}{2}(x-x^5)\vec i-(x-x^3)\vec j+y^2(1-x^2)\vec k\big]\begin{matrix} 1 \\ x^2 \end{matrix}dydx$$

$$=\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}\bigg(\dfrac{1}{2}(x-x^5)\vec i-(x-x^3)\vec j+y^2(1-x^2)\vec k\bigg)dydx$$

$$=\displaystyle\int_{0}^{1}\bigg[\big(\dfrac{1}{2}(x-x^5)y\vec i-(x-x^3)y\vec j+\dfrac{y^3}{3}(1-x^2)\vec k\big)\bigg]\begin{matrix} 1 \\ 0 \end{matrix}dx$$

$$=\displaystyle\int_{0}^{1}\bigg(\big(\dfrac{1}{2}(x-x^5)\vec i-(x-x^3)\vec j+\dfrac{1}{3}(1-x^2)\vec k\big)\bigg)dx$$

$$=[(\dfrac{1}{4}x^2-\dfrac{1}{12}x^6)\vec i-(\dfrac{1}{2}x^2-\dfrac{1}{4}x^4)\vec j+(\dfrac{1}{3}x-\dfrac{1}{9}x^3))\vec k]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=\dfrac{1}{6}\vec i-\dfrac{1}{4}\vec j+\dfrac{2}{9}\vec k$$