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

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Answer 1 · 2018-05-09T06:38:30

#A=32/3#

.

The area bounded by:

#x=y^2-4y#

#x=0#

#y=0#

#y=4#

Let's look at the graph of this area:

enter image source here

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#