Saturday

September 20, 2014

September 20, 2014

Posted by **TOMO** on Friday, August 20, 2010 at 10:58pm.

computed in two different ways using integrals.

First of all it can be computed as a sum of two integrals

They ask to use two integrals so i put f(x) from -7 to 2 which is correct

but for g(x) i put 2 to 14 for some reason 14 is wrong. I also put

f(x)=sqrt(x+7) and g(x)= (14-x)/4 and both are wrong wrong. I got

everything else correct except for these and I don't what I did wrong.

- MATH 2B Calculus -
**jai**, Saturday, August 21, 2010 at 11:57amfirst, graph the two equations (in one cartesian plane)

then get the points of intersection:

x+7=y^2 *this is the second equation

x=y^2-7

substitute this to the first:

y^2-7+4y=14

y^2+4y-21=0

(y+7)(y-3)=0

y=-7 and y=3

substitute these back to obtain corresponding values of x:

*if y=-7,

x=(-7)^2-7=42

*if y=3,

x=(3)^2-7=2

therefore, points of int are (42,-7) and (2,3)

looking at the graph, i suggest you do vertical strips (that is, dx),, divide the whole area into region 1 (left side) and region 2 (right side),, after you do this, get the boundaries.

Region 1:

for x: the boundaries are the graph of the parabola (that is x=y^2-7) and the x-coord of the first point of int (x=2)

for y: since parabola is symmetric with respect to x-axis, y is from -3 to 3.

the area of region 1 is:

integral[from -3 to 3](integral[from y^2-7 to 2] dx)dy)

*note: this is double integral since A=dxdy

Region 2:

for region 2, i suggest you do horizontal strips (that is, dy) ,,then get the boundaries:

for x: from x-coord of first point of int (x=2) to x-ccord of 2nd point of int (x=42)

for y: from the parabola (y=sqrt(x+7)) to the line (y=(14-x)/4)

the area of region 2 is:

integral[from 2 to 42](integral[from sqrt(x+7) to (14-x)/4] dy)dx)

i'll leave the integration calculation to you,, add the areas and you'll finally get the whole area.

so there,, sorry for long explanation,,

i hope i was able to help,, =)

**Answer this Question**

**Related Questions**

Cal - consider the area between the graphs x+3y=1 and x+9=y^2. this area can be ...

Cal - consider the area between the graphs x+3y=1 and x+9=y^2. this area can be ...

Calculus Area between curves - Consider the area between the graphs x+6y=8 and x...

math, calculus 2 - Consider the area between the graphs x+y=16 and x+4= (y^2). ...

Calculus Please Help4 - Consider the area between the graphs x+2y=9 and x+6=y^2...

Calculus - Consider the area between the graphs x+1y=12 and x+8=y2 . This area ...

CAL - consider the area between the graphs x+3y=1 and x+9=y^2. this area can be ...

Calculus - Integrals: When we solve for area under a curve, we must consider ...

Math - A student opened her math book and computed the sum of the numbers on two...

math, calculus - 1. Consider the region bounded by the curves y=|x^2+x-12|, x=-5...