- #1
DryRun
Gold Member
- 838
- 4
Homework Statement
http://s1.ipicture.ru/uploads/20111229/vqL4SkL9.jpg
The attempt at a solution
So, i drew the graph of y against x although i don't know initially if a>b or a<b.
http://s1.ipicture.ru/uploads/20111229/S3y3oSFS.jpg
Looking at the limits, i realized that a must be less than b. So, the shaded areas are the regions which i need to find.
Description:
For y fixed, y varies from y=0 to y=a
and x varies from x=0 to x = (a/b)√(b^2 - y^2)
To change the order of integration, d.w.r.t.y first and then d.w.r.t.x. The region is split into 2 parts.
The region A is the red section on the graph and region B is the blue section on the graph.
Region A:
For x fixed, x varies from x=0 to x=(a/b)√(b^2 - a^2)
and y varies from y=0 to y=a
Region B:
For x fixed, x varies from x = (a/b)√(b^2 - a^2) to x=a
and y varies from y=0 to y=√[b^2 - (b^2.x^2)/a^2]
At this point, i checked but not sure if my work above is correct. But I'm moving on.
Thus, after changing the order, the double integral is equal to the sum of the double integral for region A and B.
I did the calculations in my copybook and the answer has become so complicated. The answer in my notes points to something quite simple.
http://s1.ipicture.ru/uploads/20111229/vqL4SkL9.jpg
The attempt at a solution
So, i drew the graph of y against x although i don't know initially if a>b or a<b.
http://s1.ipicture.ru/uploads/20111229/S3y3oSFS.jpg
Looking at the limits, i realized that a must be less than b. So, the shaded areas are the regions which i need to find.
Description:
For y fixed, y varies from y=0 to y=a
and x varies from x=0 to x = (a/b)√(b^2 - y^2)
To change the order of integration, d.w.r.t.y first and then d.w.r.t.x. The region is split into 2 parts.
The region A is the red section on the graph and region B is the blue section on the graph.
Region A:
For x fixed, x varies from x=0 to x=(a/b)√(b^2 - a^2)
and y varies from y=0 to y=a
Region B:
For x fixed, x varies from x = (a/b)√(b^2 - a^2) to x=a
and y varies from y=0 to y=√[b^2 - (b^2.x^2)/a^2]
At this point, i checked but not sure if my work above is correct. But I'm moving on.
Thus, after changing the order, the double integral is equal to the sum of the double integral for region A and B.
I did the calculations in my copybook and the answer has become so complicated. The answer in my notes points to something quite simple.
Last edited: