Answer to Question #264741 in Calculus for JaytheCreator

Question #264741

Question 2

(i) Find the volume integral of the scalar field πœ™(π‘₯, 𝑦, 𝑧) = π‘₯^2 + 𝑦^2 + 𝑧^2 over the region 𝑉 specified by π‘₯ ∈ [0,1], 𝑦 ∈ [1,2], and 𝑧 ∈ [0,3].

(ii) Find the gradient of the scalar field 𝑓(π‘₯, 𝑦, 𝑧) = π‘₯𝑦𝑧 and evaluate it at the point (1,2,3). Hence, find the directional derivative of 𝑓 at this point in the direction of the vector (1,1,0).


1
Expert's answer
2021-11-17T17:15:09-0500

I

"X,y,z=x^2+y^2+z^2\\\\"

Region of integration is given by E(X,y,z): 0<X<1. 1<y<2. 0<y<2 0<z<3

Volume

"\\iiint(x^2+y^2+z^2)dv"

"\\intop_{x=0}^{1}\\intop_{y=1}^{2}\\intop_{z=0}^{3}(x^2+y^2+z^2)dzdydx\\\\"

"\\intop_{0}^{1}\\intop_{1}^{2}(x^2z+y^2z+\\frac{z^3}{3})dydx"

"\\intop_{0}^{1}\\intop_{1}^{2}(3x^2+3y^2z+9)dydx"

"\\intop_{0}^{1}(3x^2y+y^3+9y)^2dx\\\\\n\\intop_{0}^{1}[(6x^2+8+18)-(3x^2+1+9)]dx\\\\\n\\intop_{0}^{1}(3x^2+16)dx\\\\"

"\\int_0^1(3x^2+16)dx\\\\1+16\\\\17"


ii

gradient vector is

"<yz,xz,xy>"

required gradient vector will be

"(1,2,3)=<2\\times 3,1\\times3,1\\times2>"

"<6,3,2>"

the direction vector is given as

"=\\frac{1,1,0}{\\sqrt{1+1+0}}\\\\<\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}},0>"

the required direction vector will be

du f=f(1,2,3)

"<6,3,2>" ."<\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}},0>"

"\\frac{6}{\\sqrt{2}}+\\frac{3}{\\sqrt{2}}+0\\\\\\frac{9}{\\sqrt{2}}"

Required directionΒ "\\frac{9}{2}\\sqrt{2}"




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