Advertisements
Advertisements
Question
Find by double integration the area bounded by the parabola ๐๐=๐๐ And ๐=๐๐−๐
Solution
The parabola ๐ฆ2=4๐ฅ and the line ๐ฆ=2๐ฅ−4 intersect where (2๐ฅ−4)2=4๐ฅ
∴ `4x^2-16x+16=4x` ∴ `4x^2-20x+16=0`
∴` x^2-5x+4=0` ∴ `(x-4) (x-1)=0`
∴` x=1,4`
When x = 1, y = 2 – 4 = -2; and when x = 4, y = 8 - 4 = 4. Thus, the points of intersection are A (1, -2) and B (4, 4).
Now, consider a strip parallel to x-axis. On this strip x varies from x = y2/4 to x = (y+4)/2. The strip then moves parallel to the x-axis from y = -2 to y = 4.
∴` A= int_-2^4 int_(y^2)"^(y+4)/2 dxdy=int_-2^4 [x]_(y^2/4)^((y+4)/2) dy`
=`int_-2^4 ((y+4)/2 - y^2/4)dy`
= `1/4 int_-2^4(2y+8-y^2) dy`
=`1/4 int_-2^4(2y+8-y^2)dy`
=` 1/4 [y^2+8y-y^3/3]_-2^4`
=`1/4[(16+32-64/3)-(4-16+8/3)]`
=`1/4(60-24)`
∴` A=9``
APPEARS IN
RELATED QUESTIONS
Find the area inside the circle r=a sin๐ฝ and outside the cardioide r=a(1+cos๐ฝ )
Evaluate `int int xy(x-1)dx dy` over the region bounded by ๐๐ = ๐,๐= ๐,๐ =๐ and ๐ = ๐
Evaluate `int int(2xy^5)/sqrt(x^2y^2-y^4+1)dxdy`, where R is triangle whose vertices are (0,0),(1,1),(0,1).
Use polar co ordinates to evaluate `int int (x^2+y^2)^2/(x^2y^2)` ๐ ๐ ๐ ๐ over yhe area Common to circle `x^2+y^2=ax "and" x^2+y^2=by, a>b>0`
