Volume Integral Engineering Maths, Btech first year

Volume Integral

Let $\overrightarrow{F}$ be a Vector Point Function and let $V$ be the volume enclosed by a closed surface then the volume integral is given by: $${\iiint }_v\overrightarrow{F}.dV$$ Here $dV=dxdydz$

Sample Problem

If $\overrightarrow{F}=2z\hat{i}-x\hat{j}+y\hat{k}$, evaluate volume integral when the equation of surfaces are: $$x=0,y=0,x=2,y=4,z=x^2,z=2$$

Solution

The equations of surfaces give the limts so, limits for $x$ : 0 to 2, for $y$ : 0 to 4 and for $z$ : $x^2$ to 2 So now volume integral, $${\iiint }_v\overrightarrow{F}.dV={\int }_0^2{\int }_0^4{\int }_{x^2}^2(2z\hat{i}-x\hat{j}+y\hat{k})dxdydz$$ $$={\int }_0^2{\int }_0^4{[z^2\hat{i}-xz\hat{j}+yz\hat{k}]}_{x^2}^{2}dxdy$$ $$={\int }_0^2{\int }_0^4[4\hat{i}-2x\hat{j}+2y\hat{k}-x^4\hat{i}+x^3\hat{j}-x^{2}y\hat{k}]dxdy$$ $$={\int }_0^2{\int }_0^4[4\hat{i}-2x\hat{j}+2y\hat{k}-x^4\hat{i}+x^3\hat{j}-x^{2}y\hat{k}]dxdy$$ $$={\int }_0^2{\int }_0^4[4\hat{i}-2x\hat{j}+2y\hat{k}-x^4\hat{i}+x^3\hat{j}-x^{2}y\hat{k}]dxdy$$ $$={\int }_0^2{\int }_0^4[4\hat{i}-2x\hat{j}+2y\hat{k}-x^4\hat{i}+x^3\hat{j}-x^{2}y\hat{k}]dxdy$$ $$={\int }_0^2{[4y\hat{i}-2xy\hat{j}+y^2\hat{k}-x^{4}y\hat{i}+x^{3}y\hat{j}-\frac{x^2{y}^2}{2}\hat{k}]}_0^{4}dx$$ $$={\int }_0^2[16\hat{i}-8x\hat{j}+16\hat{k}-4x^4\hat{i}+4x^3\hat{j}-8x^2\hat{k}]dx$$ $$={[16x\hat{i}-4x^2\hat{j}+16x\hat{k}-\frac{4x^5}{5}\hat{i}+x^4\hat{j}-\frac{8x^3}{3}\hat{k}]}_0^2$$ $$=32\hat{i}-16\hat{j}+32\hat{k}-\frac{128}{5}\hat{i}+16\hat{j}-\frac{64}{3}\hat{k}$$ $$=\boxed{{\displaystyle \frac{32}{15}(3\hat{i}+5\hat{k})}}$$