Answer to Question #264741 in Calculus for JaytheCreator

Question

Question 2

(i) Find the volume integral of the scalar field 𝜙(𝑥, 𝑦, 𝑧) = 𝑥^2 + 𝑦^2 + 𝑧^2 over the region 𝑉 specified by 𝑥 ∈ [0,1], 𝑦 ∈ [1,2], and 𝑧 ∈ [0,3].

(ii) Find the gradient of the scalar field 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧 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).

Answer 1

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

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

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

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

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

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

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

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 direction $\frac{9}{2}\sqrt{2}$