This answer is not selected as best answer. This answer may not be sufficient for exam.

Your limit has been exceed. We have implemented this system because, We got difficulty on managing our servers. Please donate some amount to remove this limit.

**Quota:** 0 / 30

Solution:

Given

Given Curves are

y = x^{2} — (i)

y = 1 — (ii)

y = 2 — (iii)

Clearly (i) is a parabola that has a vertices at (0, 0) and the line of symmetry is x = 0 with y ≥ 0.

Also, (ii) and (iii) are the straight lines that are parallel to x-axis. With these information, the sketch in figure, in which the bounded region by (i), (ii), (iii) is the shaded portion.

Clearly the region has two symmetrical parts I and II. From figure, part – I, y moves from 1 to 2.

Now, the area of the bounded region is

\(\)

\(Area = 2\left | \int_{y = 1}^{2} x dy \right |\)

= \(2\left | \int_{1}^{2} \sqrt{y} dx \right | \)

= \(2\left [ \frac{y^{\frac{3}{2}}}{\frac{3}{2}} \right ]^2_1\)

= \(\frac{4}{3} (2^{\frac{3}{2}} – 1^{\frac{3}{2}})\)

= 3.77 sq. units

Thus, the area of the region is 3.77 sq. units

If you found any type of error on the answer then please mention on the comment or submit your new answer.

Click here to submit your answer.

HAMROCSIT.COM

## Discussion