Chapter 5: Q. 6 (page 466)

Calculate the surface area of the solid of revolution obtained by revolving the region between $f (x) = 2 x^{3}$ and the $x$ -axis on $[0, 3]$ around the $x$ -axis.

Short Answer

Expert verified

The surface area of the solid of revolution obtained by revolving the region between $f (x) = 2 x^{3}$ and the $x$ -axis on $[0, 3]$ around the $x$ -axis is, $1526.81$ (approximately) square units.

Step by step solution

Step 1. Given information

The region between $f (x) = 2 x^{3}$ and the x-axis on $[0, 3]$ .

Step 2. The formula for the surface area is,

Surface area $= 2 π \int_{a}^{b} f (x) \sqrt{1 + {(f^{'} (x))}^{2}} dx$ .

Let $f (x) = 2 x^{3}, f^{'} (x) = 6 x^{2}$ .

Substitute the known values in the formula.

Surface area $= 2 π \int_{0}^{3} 2 x^{3} \sqrt{1 + {(6 x^{2})}^{2}} dx$

role="math" localid="1648890749974" $= 2 π \int_{0}^{3} 2 x^{3} \sqrt{1 + 36 x^{4}} dx$

Let $u = 1 + 36 x^{4}, du = 144 x^{3}$ .

role="math" localid="1648892010258" $2 π \int_{x = 0}^{x = 3} 2 x^{3} \sqrt{1 + 36 x^{4}} dx = 2 π (\frac{2}{144}) \int_{x = 0}^{x = 3} \sqrt{u} du = (\frac{π}{36}) {[\frac{2}{3} u^{\frac{3}{2}}]}_{0}^{3} = (\frac{π}{36}) (\frac{2}{3}) {[{(1 + 36 x^{4})}^{\frac{3}{2}}]}_{0}^{3} = (\frac{π}{54}) {[(1 + 36 x^{6})]}_{0}^{3} = (\frac{π}{54}) [(1 + 36 {(3)}^{6}) - (1 + 36 {(0)}^{6})] = (\frac{π}{54}) [1 + 26, 244 - 1] = 486 π ≅ 1526.81$

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Calculate the surface area of the solid of revolution obtained by revolving the region between $f (x) = 2 x^{3}$ and the $x$ -axis on $[0, 3]$ around the $x$ -axis.

Short Answer

Step by step solution

Step 1. Given information

Step 2. The formula for the surface area is,

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Calculate the surface area of the solid of revolution obtained by revolving the region between fx=2x3and the x-axis on 0,3around the x-axis.

Short Answer

Step by step solution

Step 1. Given information

Step 2. The formula for the surface area is,

One App. One Place for Learning.

Most popular questions from this chapter

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Calculate the surface area of the solid of revolution obtained by revolving the region between $f (x) = 2 x^{3}$ and the $x$ -axis on $[0, 3]$ around the $x$ -axis.