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Double integral to find area of a portion.

  • Thread starter raoulduke1
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  • #1
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Homework Statement



the question is 3(b) on the attached pdf.




Homework Equations





The Attempt at a Solution



I could only get as far as the filling in the equation.
How do they change it to one integral.?
And also where did they get them substitutions from?

Any help would be greatly appeciated.
 

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Answers and Replies

  • #2
DryRun
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First, you should make the sketch to understand how to project the surface. You haven't shown any work in your post.

The substitution used in the integral is due to the presence of ##\sqrt{8-y^2}=\sqrt{(\sqrt{8})^2-y^2}##
 
  • #3
Simon Bridge
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I could only get as far as the filling in the equation.
Which would be formula (f) from the table at the end. Did you manage to sketch the region?
How do they change it to one integral.?
They are integrating over x and y ... they just did the integration over x first.
And also where did they get them substitutions from?
A physicist would use the sketch for part (a) to change coordinates soas to exploit the symmetry. Mathematicians are more formal... what they have described is a parameterization (in effect changing to cylindrical coordinates).

The substitutions are basically trig substitutions.
You know that if you have an integral like [tex]\int \frac{dx}{\sqrt{a-bx^2}}[/tex]... then you want to try a substitution of form [tex]x=\sqrt{\frac{a}{b}}\sin(t)[/tex].
 
  • #4
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Which would be formula (f) from the table at the end. Did you manage to sketch the region?They are integrating over x and y ... they just did the integration over x first.
A physicist would use the sketch for part (a) to change coordinates soas to exploit the symmetry. Mathematicians are more formal... what they have described is a parameterization (in effect changing to cylindrical coordinates).

The substitutions are basically trig substitutions.
You know that if you have an integral like [tex]\int \frac{dx}{\sqrt{a-bx^2}}[/tex]... then you want to try a substitution of form [tex]x=\sqrt{\frac{a}{b}}\sin(t)[/tex].
Yes I sketched it.

so its basically a rectangle split into 4?
does that 4 before the integral mean they are integrating a quarter of the shape and multiplying it by 4?
 
  • #5
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also how did the integral value change at the end from (2 and 0) to (pi/2 and 0)
 
  • #6
Simon Bridge
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so its basically a rectangle split into 4?
No.

The surface is a cylinder, radius 2√2, aligned along the x axis.
The region is a 2x4 rectangle, centered at the origin, lying in the x-y plane, long side aligned to the y axis.

If your picture does not look like a section of a cylinder, then it is no wonder you are having trouble. In fact - if you had the right diagram, you could have worked out the area just using that and basic High School geometry: no calculus needed.

The 4 in front is partly the result of integrating the dx part. The rest is from using the symmetry. (the -y part is the same as the +y)

The limits of integration change with the substitution: the old limits were for y, the new limits are for t ... complete: when y=-2, t=?
 

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