Question

The area of the region
$\{ ( \rm{x , }y)\ : xy \leqslant 8,1 \leqslant y \leqslant {x}^{2} \}$
is ​

Answer 1

Topic: Area Under the Curve:

Intially we need to draw the graph of the given curved inorder to integrate and find the values

1.) xy = 8 (this is the graph of a rectangular hyperbola)

2.) then we are provided with y = x² (an upward facing parabola touching the x axis at the origin)

3.) we are also given y = 1 i.e a straight line cutting y axis at 1

the required curve by plotting these graphs is shown in the attachment.

NOTE: IT IS EXTREMELY IMPORTANT TO MARK THE INTERSECTING POINTS OF THE GRAPH

you can find the intersecting points between the parabola and the rectangular hyperbola by equating the curves as:

• 8/x = x²
• x = 2

intersecting point between the straight line and hyperbola:

• 1 = 8/x
• x = 8

intersecting point between parabola and stright line

• x² = 1
• x = 1

Now, we have to integrate the curve w.r.t y since the initial and final points of all the graphs is same along y axis

Solution is provided in the attachment 2 :)

Answer 2

xy ≤8

1 ≤ y ≤ x²

x².x = 8

x = 2

Require Area

$\sf = ∫_{1}^{2} ( \frac{8}{y} - \sqrt{y} )dy =$

$[ \sf 8lny - \frac{ {y}^{3/2} }{3/2} ]_{1}^{2}$

$\sf = 8 \: lny4 - \frac{2}{3} . 8 - 0 + \frac{2}{3} = 16ln2 - \frac{14}{3} .$

${\mathtt {\color {white}{\colorbox{#FF0000} {Hope it is helpful you!!}}}}$