Sketch the region enclosed by the curves x=y^2-4y, x=0, y=0, y=4 and find its area?

1 Answer
May 9, 2018

A=32/3

Explanation:

.

The area bounded by:

x=y^2-4y

x=0

y=0

y=4

Let's look at the graph of this area:

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This area is between the sideways parabola shown in purple and the y-axis. Let's find out the y-intercepts.

x=0, :. y=0 and 4

We will integrate the sideways parabola function with respect to y (along the y-axis) and evaluate it between y=0 and 4:

A=int_0^4(y^2-4y)dy=(1/3y^3-2y^2)_0^4=1/3(4)^3-2(4)^2-(1/3(0)^3-2(0)^2)=64/3-32=abs(-32/3)=32/3