If the line x=3 divides the area bounded between the curves y^2=12x and x^2=12y into two parts then the ratio of their areas is
4 answers
what do you mean by "their areas"?
take a look at the graphs
http://www.wolframalpha.com/input/?i=y%5E2%3D12x+and+x%5E2%3D12y
clearly they intersect at (0,0) and (12,12)
height of region = 2√3 x^(1/2) - (1/12)x^2
area from 0 to 3
= ∫ (2√3 x^(1/2) - (1/12)x^2) dx from 0 to 3
= [ 4√3/3 x^(3/2) - (1/36)x^3 ] from 0 to 3
= 4√3/3 (3√3) - 27/36 - 0
= 12 - 3/4 = 45/4
area from 0 to 12
= ∫ (2√3 x^(1/2) - (1/12)x^2) dx from 0 to 12
= [ 4√3/3 x^(3/2) - (1/36)x^3 ] from 0 to 12
= 4√3/3 (12)^(3/2) - (1/36)(1728)
= 96 - 48 = 48
so area from 3 to 12 = 48 - 45/4 = 147/4
ratio of areas = (45/4) : (147/4)
= 45 : 147
= 15 : 49
http://www.wolframalpha.com/input/?i=y%5E2%3D12x+and+x%5E2%3D12y
clearly they intersect at (0,0) and (12,12)
height of region = 2√3 x^(1/2) - (1/12)x^2
area from 0 to 3
= ∫ (2√3 x^(1/2) - (1/12)x^2) dx from 0 to 3
= [ 4√3/3 x^(3/2) - (1/36)x^3 ] from 0 to 3
= 4√3/3 (3√3) - 27/36 - 0
= 12 - 3/4 = 45/4
area from 0 to 12
= ∫ (2√3 x^(1/2) - (1/12)x^2) dx from 0 to 12
= [ 4√3/3 x^(3/2) - (1/36)x^3 ] from 0 to 12
= 4√3/3 (12)^(3/2) - (1/36)(1728)
= 96 - 48 = 48
so area from 3 to 12 = 48 - 45/4 = 147/4
ratio of areas = (45/4) : (147/4)
= 45 : 147
= 15 : 49
Stupid me. I read the question as "two equal parts" and wondered what the fuss was about. Clearly the ratio would be 1:1
Nice catch, Reiny.
Nice catch, Reiny.
Pls also show the diagram
1 answer