Question

The area of the region $$R = \left\{ {\left( {x,\,y} \right):\left| x \right| \leqslant \left| y \right|{\text{ and }}{x^2} + {y^2} \leqslant 1} \right\}$$        is :

A. $$\frac{{3\pi }}{8}{\text{ sq}}{\text{. unit}}$$
B. $$\frac{{5\pi }}{8}{\text{ sq}}{\text{. unit}}$$
C. $$\frac{\pi }{2}{\text{ sq}}{\text{. unit}}$$  
D. $$\frac{\pi }{8}{\text{ sq}}{\text{. unit}}$$
Answer :   $$\frac{\pi }{2}{\text{ sq}}{\text{. unit}}$$
Solution :
Required area $$=$$ area of the shaded region
$$ = 4$$  (area of the shaded region in first quadrant
$$\eqalign{ & = 4\int_0^{\frac{1}{{\sqrt 2 }}} {\left( {{y_1} - {y_2}} \right)} dx \cr & = 4\int_0^{\frac{1}{{\sqrt 2 }}} {\left( {\sqrt {1 - {x^2}} - x} \right)dx} \cr & = 4\left[ {\frac{1}{2} \times \sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}x - \frac{{{x^2}}}{2}} \right]_0^{\frac{1}{{\sqrt 2 }}} \cr & = \frac{\pi }{2}{\text{ sq}}{\text{. unit}} \cr} $$
Application of Integration mcq solution image

Releted MCQ Question on
Calculus >> Application of Integration

Releted Question 1

The area bounded by the curves $$y = f\left( x \right),$$   the $$x$$-axis and the ordinates $$x = 1$$  and $$x = b$$  is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$     Then $$f\left( x \right)$$  is-

A. $$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
B. $$\sin \,\left( {3x + 4} \right)$$
C. $$\sin \,\left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
D. none of these
Releted Question 2

The area bounded by the curves $$y = \left| x \right| - 1$$   and $$y = - \left| x \right| + 1$$   is-

A. $$1$$
B. $$2$$
C. $$2\sqrt 2 $$
D. $$4$$
Releted Question 3

The area bounded by the curves $$y = \sqrt x ,\,2y + 3 = x$$    and $$x$$-axis in the 1st quadrant is-

A. $$9$$
B. $$\frac{{27}}{4}$$
C. $$36$$
D. $$18$$
Releted Question 4

The area enclosed between the curves $$y = a{x^2}$$   and $$x = a{y^2}\left( {a > 0} \right)$$    is 1 sq. unit, then the value of $$a$$ is-

A. $$\frac{1}{{\sqrt 3 }}$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{1}{3}$$

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