Find the area bounded by the curve x = y3 - 2y2 - 3y and the y-axis.

Question

Calculus question that i am having trouble with

Answer 1

The area will be exactly the same as bounded by the curve y = x3 - 2x2 - 3x and the x-axis

 

I just switched x and y. Why did I do that? Because, the way we draw x-axis as a horizontal line and y-axis as a vertical line, it's easier to visualize y = f(x), rather than x = g(y)

 

So, let's find the area bounded by y = x3 - 2x2 - 3x and the x-axis.

 

Area = ∫ f(x) dx and we sum over each section of the curve where it is bounded by the x-axis.

 

For this, we need to know the intersection points of the curve f(x) with x-axis.

f(x) = 0

x3 - 2x2 - 3x = 0

x(x2-2x-3) = 0

x(x2-3x+x-3) = 0

x(x-3)(x+1) = 0

x = -1, 0 and 3

 

so, the curve intersects x-axis at three points: -1, 0 and 3

 

The bounded part of the curve with the x-axis will be from -1 to 0 and then from 0 to 3. The portions of the curve for x =(-∞, -1) and (3, +∞) will NOT be bounded by the x-axis.

 

So we apply area under the curve formula two times, first from (-1, 0) and then from (0,3). Whatever numbers we get, we take the positive, because area is always positive.

 

Total area = |∫ (x3 - 2x2 - 3x) dx, (-1, 0)| + |∫ (x3 - 2x2 - 3x) dx, (0, 3)|

 

Note the absolute value sign, which indicates that we take the positive value for the area.

 

I am leaving above integrals for you to work on. Final answer I am getting is 71/6 = 11.83

 

Hope this helps!

Answer 2

The area bounded by a curve and the y-axis between limits y=a and y=b

 

is integral from a to b (xdy).  So first we must find the limits and graphing the curve should help.

 

x=y^3-2y^2-3y=y(y-3)(y+1)

 

So the curve intersects the x-axis at y=-1,0,3

 

dx/dy=3y^2-4y-3=0 which has roots at 2/3+/-sqrt(16+4*3*3)/6=(2+/-sqrt(13))/3 lying

in the intervals (-1,0) and (0,3) respectively on the y-axis.  At these points the curve's slope is parallel to y-axis.

 

As y goes to infinity so does x...as y goes to -infinity so does x.

 

The curve is bounded in the region (-1,3) with a positive area -1<y>0 and negative 0<y<3

and unbounded elsewhere with a infinitely positive area y>3 and negative y<-1

 

integral from y=-1 to 0 xdy=( y^4)/4-2(y^3)/3-3(y^2)/2 evaluated at these limits

=-1/4-2/3+3/2=7/12

 

integral from 0 to 3......

=81/4-18-27/2=-45/4

 

integral -inf to -1.... +integral 3 to inf=-integral -1 to -inf ......+integral 3 to inf where in the limit

the y^4 term disappears but the y^3 term dominates and net this area is infinitely negative.

 

So if we require area to be positive and bounded you are looking for abs(7/12)+ abs(-45/4)=11 5/6