1) x² + y² = 16 from x = 2 to x = 4
f ( x ) = = 16 − y 2 f(x)==\sqrt{16-y^2} f ( x ) == 16 − y 2
The surface area of the curve is given byS = 2 π ∫ a b f ( x ) ( f ′ ( x ) ) 2 + 1 d x S = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x S = 2 π ∫ a b f ( x ) ( f ′ ( x ) ) 2 + 1 d x
First, find the derivative
f ′ ( x ) = ( 16 − x 2 ) ′ = − x 16 − x 2 f '\left(x\right)=\left(\sqrt{16 - x^{2}}\right)'=- \frac{x}{\sqrt{16 - x^{2}}} f ′ ( x ) = ( 16 − x 2 ) ′ = − 16 − x 2 x
Finally, calculate the integral
S = ∫ 2 4 2 π 16 − x 2 ( − x 16 − x 2 ) 2 + 1 d x S = \int_{2}^{4} 2 \pi \sqrt{16 - x^{2}} \sqrt{\left(- \frac{x}{\sqrt{16 - x^{2}}}\right)^{2} + 1} d x S = ∫ 2 4 2 π 16 − x 2 ( − 16 − x 2 x ) 2 + 1 d x
= ∫ 2 4 8 π − 1 x 2 − 16 16 − x 2 d x =\int_{2}^{4} 8 \pi \sqrt{- \frac{1}{x^{2} - 16}} \sqrt{16 - x^{2}} d x = ∫ 2 4 8 π − x 2 − 16 1 16 − x 2 d x
=∫ 2 4 ( 8 π − 1 x 2 − 16 16 − x 2 ) d x = ∫ 2 4 ( 8 π ) d x \int_{2}^{4}\left( 8 \pi \sqrt{- \frac{1}{x^{2} - 16}} \sqrt{16 - x^{2}} \right)dx=\int_{2}^{4}\left( 8 \pi \right)dx ∫ 2 4 ( 8 π − x 2 − 16 1 16 − x 2 ) d x = ∫ 2 4 ( 8 π ) d x
= ∫ 2 4 ( 8 π ) d x = ( 8 π x ) ∣ ( x = 4 ) − ( 8 π x ) ∣ ( x = 2 ) = 16 π =\int_{2}^{4}\left( 8 \pi \right)dx=\left(8 \pi x\right)|_{\left(x=4\right)}-\left(8 \pi x\right)|_{\left(x=2\right)}=16 \pi = ∫ 2 4 ( 8 π ) d x = ( 8 π x ) ∣ ( x = 4 ) − ( 8 π x ) ∣ ( x = 2 ) = 16 π
= 16 π ≈ 50.2654824574367 =16 \pi\\ \approx \boxed{50.2654824574367} = 16 π ≈ 50.2654824574367
2) y² = 12x from x = 0 to x = 3
f ( x ) = 2 3 x f\left(x\right)=2 \sqrt{3} \sqrt{x} f ( x ) = 2 3 x
he surface area of the curve is given byS = 2 π ∫ a b f ( x ) ( f ′ ( x ) ) 2 + 1 d x S = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x S = 2 π ∫ a b f ( x ) ( f ′ ( x ) ) 2 + 1 d x
f ′ ( x ) = ( 2 3 x ) ′ = 3 x f '\left(x\right)=\left(2 \sqrt{3} \sqrt{x}\right)'=\frac{\sqrt{3}}{\sqrt{x}} f ′ ( x ) = ( 2 3 x ) ′ = x 3
calculate the integral
S = ∫ 0 3 2 π 2 3 x ( 3 x ) 2 + 1 d x = ∫ 0 3 4 3 π x x + 3 x d x S = \int_{0}^{3} 2 \pi 2 \sqrt{3} \sqrt{x} \sqrt{\left(\frac{\sqrt{3}}{\sqrt{x}}\right)^{2} + 1} d x \\=\int_{0}^{3} 4 \sqrt{3} \pi \sqrt{x} \sqrt{\frac{x + 3}{x}} d x S = ∫ 0 3 2 π 2 3 x ( x 3 ) 2 + 1 d x = ∫ 0 3 4 3 π x x x + 3 d x
= ∫ 0 3 ( 4 π 3 x + 9 ) d x =\int_{0}^{3}\left( 4 \pi \sqrt{3 x + 9} \right)dx = ∫ 0 3 ( 4 π 3 x + 9 ) d x
∫ 0 3 ( 4 π 3 x + 9 ) d x = ( 8 3 π ( x + 3 ) 3 2 3 ) ∣ ( x = 3 ) − ( 8 3 π ( x + 3 ) 3 2 3 ) ∣ ( x = 0 ) = − 24 π + 48 2 π \int_{0}^{3}\left( 4 \pi \sqrt{3 x + 9} \right)dx=\left(\frac{8 \sqrt{3} \pi \left(x + 3\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=3\right)}-\left(\frac{8 \sqrt{3} \pi \left(x + 3\right)^{\frac{3}{2}}}{3}\right)|_{\left(x=0\right)}=- 24 \pi + 48 \sqrt{2} \pi ∫ 0 3 ( 4 π 3 x + 9 ) d x = ( 3 8 3 π ( x + 3 ) 2 3 ) ∣ ( x = 3 ) − ( 3 8 3 π ( x + 3 ) 2 3 ) ∣ ( x = 0 ) = − 24 π + 48 2 π
= − 24 π + 48 2 π ≈ 137.860157345447 =- 24 \pi + 48 \sqrt{2} \pi\\\approx \boxed{137.860157345447} = − 24 π + 48 2 π ≈ 137.860157345447
3) y = x³ from x = 0 to x = 1
f(x)=x3
S = 2 π ∫ a b f ( x ) ( f ′ ( x ) ) 2 + 1 d x S = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x S = 2 π ∫ a b f ( x ) ( f ′ ( x ) ) 2 + 1 d x
f ′ ( x ) = ( x 3 ) ′ = 3 x 2 f '\left(x\right)=\left(x^{3}\right)'=3 x^{2} f ′ ( x ) = ( x 3 ) ′ = 3 x 2
S = ∫ 0 1 2 π x 3 ( 3 x 2 ) 2 + 1 d x = ∫ 0 1 2 π x 3 9 x 4 + 1 d x S = \int_{0}^{1} 2 \pi x^{3} \sqrt{\left(3 x^{2}\right)^{2} + 1} d x=\int_{0}^{1} 2 \pi x^{3} \sqrt{9 x^{4} + 1} d x S = ∫ 0 1 2 π x 3 ( 3 x 2 ) 2 + 1 d x = ∫ 0 1 2 π x 3 9 x 4 + 1 d x
∫ 0 1 ( 2 π x 3 9 x 4 + 1 ) d x = ( π ( 9 x 4 + 1 ) 3 2 27 ) ∣ ( x = 1 ) − ( π ( 9 x 4 + 1 ) 3 2 27 ) ∣ ( x = 0 ) = − π 27 + 10 10 π 27 \int_{0}^{1}\left( 2 \pi x^{3} \sqrt{9 x^{4} + 1} \right)dx \\=\left(\frac{\pi \left(9 x^{4} + 1\right)^{\frac{3}{2}}}{27}\right)|_{\left(x=1\right)}-\left(\frac{\pi \left(9 x^{4} + 1\right)^{\frac{3}{2}}}{27}\right)|_{\left(x=0\right)}\\=- \frac{\pi}{27} + \frac{10 \sqrt{10} \pi}{27} ∫ 0 1 ( 2 π x 3 9 x 4 + 1 ) d x = ( 27 π ( 9 x 4 + 1 ) 2 3 ) ∣ ( x = 1 ) − ( 27 π ( 9 x 4 + 1 ) 2 3 ) ∣ ( x = 0 ) = − 27 π + 27 10 10 π
≈ 3.56312185201375 \approx \boxed{3.56312185201375} ≈ 3.56312185201375
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