- #1
jaychay
- 58
- 0
Can you check it for me please that I have done it right or not ?
Thank you in advance.
Find the volume by using shell and disk method
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SUMMARY
The discussion focuses on calculating the volume of a solid of revolution using both the shell and disk methods. The correct equations for the curves are established as \( R = 2 - \cos{x} \) and \( r = 2 - \sqrt{4 - x^2} \). Additionally, the upper branch of the parabola is defined by \( y = \sqrt{x - 1} + 1 \), and an alternative approach involves translating the graphs to place the parabola vertex at the origin, represented by \( x = y^2 \) and the line by \( y = 2 - x \), with rotation about the line \( y = -1 \).
PREREQUISITES- Understanding of integral calculus, specifically volume calculations using the shell and disk methods.
- Familiarity with curve equations and transformations in Cartesian coordinates.
- Knowledge of trigonometric functions and their properties.
- Ability to manipulate and interpret mathematical expressions and graphs.
- Study the shell method for volume calculations in detail.
- Explore the disk method and its applications in solid geometry.
- Learn about graph transformations and their effects on equations.
- Investigate the properties of trigonometric functions in volume problems.
Students and educators in calculus, mathematicians focusing on solid geometry, and anyone interested in mastering volume calculations using advanced integration techniques.
Can you check it for me please that I have done it right or not ?
Thank you in advance.
- #2
skeeter
- 1,103
- 1
1st one, your integrand is reversed.
$R= 2-\cos{x}$,
$r = 2-\sqrt{4-x^2}$
2nd ... note the curve equation is incorrect.
should be $x=(y-1)^2+1 \implies y = \sqrt{x-1}+1$ for the upper branch of the parabola
alternatively, you could translate both graphs such that the parabola vertex is at the origin, $x=y^2$, with the line having the equation $y = 2-x$ and rotating the shaded region about $y=-1$
$R= 2-\cos{x}$,
$r = 2-\sqrt{4-x^2}$
2nd ... note the curve equation is incorrect.
should be $x=(y-1)^2+1 \implies y = \sqrt{x-1}+1$ for the upper branch of the parabola
alternatively, you could translate both graphs such that the parabola vertex is at the origin, $x=y^2$, with the line having the equation $y = 2-x$ and rotating the shaded region about $y=-1$
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