Find the surface area of the torus shown below, with radii r and R.

The main objective of this question is to find the surface area of the given torus with the radii represented by r and R.

This question uses the concept of the torus. A torus is basically the surface revolution generated as a result of rotating the circle in the three-dimensional space.

Expert Answer

In this question, we will aim to find the surface area of the torus whose radius of the tube is r and the distance to the center is R.

We know that torus generated as a result of rotating circle is:

$$(x–R{)}^2+y^2=r^2,R>r>0$$

The top half is:

$$f(x)=(r^2–(x–R^2{)}^{\frac{1}{2}},R–r\le x\le R+r$$

Thus:

$$x\in [x_0,x_0+\mathrm{\Delta }x]$$

$$\mathrm{\Delta }s=\sqrt{(\mathrm{\Delta }x{)}^2+(f(x_o+\mathrm{\Delta }x)–f(x_o){)}^2}$$

$$ds=\sqrt{1+(f^{\prime }(x){)}^2}$$

Then:

$$dA=2\pi xds=2\pi x\sqrt{1+(f^{\prime }(x){)}^2}dx$$

$$f^{\prime }(x)=\frac{1}{2}(r^2–(x–R{)}^2{)}^{\frac{1}{2}}2(R–x)$$

$$=\frac{R–x}{f(x)}$$

$$=\sqrt{1+(f^{\prime }(x){)}^2}=\frac{x}{f(x)}$$

Thus:

$$2A=4{\pi }^{2}Rr$$

Numerical Answer:

The surface area of the torus is $4{\pi }^{2}Rr$.

Example

Find the surface area of the torus whose radii are r and r.

In this question, we will aim to find the surface area of the torus whose radius of the tube is r and the distance to the center r.

Torus generated as a result of rotating circle is:

$$(x–r{)}^2+y^2=r^2,r>r>0$$

The top half is:

$$f(x)=(r^2–(x–r^2{)}^{\frac{1}{2}},r–r\le x\le r+r$$

Thus by simplifying, we get:

$$x\in [x_0,x_0+\mathrm{\Delta }x]$$

$$\mathrm{\Delta }s=\sqrt{(\mathrm{\Delta }x{)}^2+(f(x_o+\mathrm{\Delta }x)–f(x_o){)}^2}$$

$$ds=\sqrt{1+(f^{\prime }(x){)}^2}$$

Then:

$$dA=2\pi xds=2\pi x\sqrt{1+(f^{\prime }(x){)}^2}dx$$

$$f^{\prime }(x)=\frac{1}{2}(r^2–(x–R{)}^2{)}^{\frac{1}{2}}2(r–x)$$

$$=\frac{r–x}{f(x)}$$

$$=\sqrt{1+(f^{\prime }(x){)}^2}=\frac{x}{f(x)}$$

By simplifying we get the surface area of the torus as:

$$2A=4{\pi }^{2}rr$$

Hence, the surface area of the torus is $space4{\pi }^{2}rr$.

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