Question #186710

Determine the surface area of the solid obtained by rotating:


1) x² + y² = 16 from x = 2 to x = 4


2) y² = 12x from x = 0 to x = 3


3) y = x³ from x = 0 to x = 1


1
Expert's answer
2021-05-12T03:47:43-0400

1) x² + y² = 16 from x = 2 to x = 4


f(x)==16y2f(x)==\sqrt{16-y^2}


The surface area of the curve is given byS=2πabf(x)(f(x))2+1dxS = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x


First, find the derivative


f(x)=(16x2)=x16x2f '\left(x\right)=\left(\sqrt{16 - x^{2}}\right)'=- \frac{x}{\sqrt{16 - x^{2}}}


Finally, calculate the integral


S=242π16x2(x16x2)2+1dxS = \int_{2}^{4} 2 \pi \sqrt{16 - x^{2}} \sqrt{\left(- \frac{x}{\sqrt{16 - x^{2}}}\right)^{2} + 1} d x


=248π1x21616x2dx=\int_{2}^{4} 8 \pi \sqrt{- \frac{1}{x^{2} - 16}} \sqrt{16 - x^{2}} d x


=24(8π1x21616x2)dx=24(8π)dx\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

=24(8π)dx=(8πx)(x=4)(8πx)(x=2)=16π=\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


=16π50.2654824574367=16 \pi\\ \approx \boxed{50.2654824574367}



2) y² = 12x from x = 0 to x = 3


f(x)=23xf\left(x\right)=2 \sqrt{3} \sqrt{x}

he surface area of the curve is given byS=2πabf(x)(f(x))2+1dxS = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x


f(x)=(23x)=3xf '\left(x\right)=\left(2 \sqrt{3} \sqrt{x}\right)'=\frac{\sqrt{3}}{\sqrt{x}}


calculate the integral

S=032π23x(3x)2+1dx=0343πxx+3xdxS = \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

=03(4π3x+9)dx=\int_{0}^{3}\left( 4 \pi \sqrt{3 x + 9} \right)dx


03(4π3x+9)dx=(83π(x+3)323)(x=3)(83π(x+3)323)(x=0)=24π+482π\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


=24π+482π137.860157345447=- 24 \pi + 48 \sqrt{2} \pi\\\approx \boxed{137.860157345447}



3) y = x³ from x = 0 to x = 1



f(x)=x3

S=2πabf(x)(f(x))2+1dxS = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x


f(x)=(x3)=3x2f '\left(x\right)=\left(x^{3}\right)'=3 x^{2}



S=012πx3(3x2)2+1dx=012πx39x4+1dxS = \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

01(2πx39x4+1)dx=(π(9x4+1)3227)(x=1)(π(9x4+1)3227)(x=0)=π27+1010π27\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}


3.56312185201375\approx \boxed{3.56312185201375}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS