Volume Integral
Let \(\vec{F}\) be a Vector Point Function and let \(V\) be the volume enclosed by a closed surface then the volume integral is given by: \[ \iiint_v \vec{F}.dV \] Here \(dV = dxdydz \)
Sample Problem
If \(\hspace{5pt} \vec{F} = 2z\hat{i} - x\hat{j} + y\hat{k} \), 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, \[ \iiint_v \vec{F}.dV = \int_0^2 \int_0^4 \int_{x^2}^2 (2z\hat{i} - x\hat{j} + y\hat{k})dxdydz \] \[ = \int_0^2 \int_0^4 \left[z^2\hat{i} - xz\hat{j} + yz\hat{k} \right]_{x^2}^2 dxdy \] \[ = \int_0^2 \int_0^4 \left[4\hat{i} - 2x\hat{j} + 2y\hat{k} - x^4\hat{i} + x^3\hat{j} - x^2y\hat{k} \right]dxdy \] \[ = \int_0^2 \int_0^4 \left[4\hat{i} - 2x\hat{j} + 2y\hat{k} - x^4\hat{i} + x^3\hat{j} - x^2y\hat{k} \right]dxdy \] \[ = \int_0^2 \int_0^4 \left[4\hat{i} - 2x\hat{j} + 2y\hat{k} - x^4\hat{i} + x^3\hat{j} - x^2y\hat{k} \right]dxdy \] \[ = \int_0^2 \int_0^4 \left[4\hat{i} - 2x\hat{j} + 2y\hat{k} - x^4\hat{i} + x^3\hat{j} - x^2y\hat{k} \right]dxdy \] \[ = \int_0^2 \left[4y\hat{i} - 2xy\hat{j} + y^2\hat{k} - x^4y\hat{i} + x^3y\hat{j} - \frac{x^2y^2}{2}\hat{k} \right]_0^4 dx \] \[ = \int_0^2 \left[16\hat{i} - 8x\hat{j} + 16\hat{k} - 4x^4\hat{i} + 4x^3\hat{j} - 8x^2 \hat{k} \right]dx \] \[ = \left[16x\hat{i} - 4x^2\hat{j} + 16x\hat{k} - \frac{4x^5}{5}\hat{i} + x^4\hat{j} - \frac{8x^3}{3} \hat{k} \right]^2_0 \] \[ = 32\hat{i} - 16\hat{j} + 32\hat{k} - \frac{128}{5}\hat{i} + 16\hat{j} - \frac{64}{3} \hat{k} \] \[ = \boxed{\frac{32}{15}(3\hat{i} + 5\hat{k})} \]
