How are the surface area and volume of the cube different?

2 answers:

7 0

The surface area is the area of one of the cube's faces. The volume, however is how much of something could fit inside of the cube.

Oxana [17]3 years ago

5 0

The surface area is the overall area of the outside of the cube; the area of all six sides are added together to get the surface area. While the volume (l•w•h) is the measure of the inside of the cube - how much matter it could possibly hold.

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Solve the problem, calculate the line integral of f along h

Over [174]

The curve $\mathcal H$ is parameterized by

$\begin{cases}X(t)=R\cos t\\Y(t)=R\sin t\\Z(t)=Pt\end{cases}$

so in the line integral, we have

$\displaystyle\int_{\mathcal H}f(x,y,z)\,\mathrm ds=\int_{t=0}^{t=2\pi}f(X(t),Y(t),Z(t))\sqrt{\left(\frac{\mathrm dX}{\mathrm dt}\right)^2+\left(\frac{\mathrm dY}{\mathrm dt}\right)^2+\left(\frac{\mathrm dZ}{\mathrm dt}\right)^2}\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}Y(t)^2\sqrt{(-R\sin t)^2+(R\cos t)^2+P^2}\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}R^2\sin^2t\sqrt{R^2+P^2}\,\mathrm dt$
$=\displaystyle\frac{R^2\sqrt{R^2+P^2}}2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt$
$=\pi R^2\sqrt{R^2+P^2}$

You are mistaken in thinking that the gradient theorem applies here. Recall that for a scalar function $f:\mathbb R^n\to\mathbb R$ , we have gradient $\nabla f:\mathbb R^n\to\mathbb R^n$ . The theorem itself then says that the line integral of $\nabla f(x,y,z)=\mathbf f(x,y,z)$ along a curve $C$ parameterized by $\mathbf r(t)$ , where $a\le t\le b$ , is given by

$\displaystyle\int_C\mathbf f(x,y,z)\,\mathrm d\mathbf r=f(\mathbf r(b))-f(\mathbf r(a))$

Specifically, in order for this theorem to even be considered in the first place, we would need to be integrating with respect to a vector field.

But this isn't the case: we're integrating $f(x,y,z)=y^2$ , a scalar function.