Question

$\huge{\red{\star}} {\blue{\huge{\mathfrak{Question}}}} \huge{\red{\star}}$



The area enclosed by the curves

y = sin(x) + cos(x) and

y = | cos(x) - sin(x) |

Over the interval [0°, 180°] is



No Spam!! ​

Answer 1

Answer:

Area = 2 areal units.

Volume of the solid of revolution = π22 cubic units.

Explanation:

Area = ∫ydx=∫sinxdx, between the limits x=0andx=π

= [−cosx], between the limits

=[−cosπ+cos0]=[1+1]=2 areal units.

Volume = π∫y2dx=π∫sin2xdx, between the limits x=0andx=π

=π2∫(1−cos2x)dx, between the limits x=0andx=π

=π2[x−sin2x2], between the limits

=π2[(π−0)−(0−0)]=π22 cubic units/

Answer 2

cosx>sinx,∀xϵ(0,

4

π

​

), and cosx<sinx,∀xϵ(

4

π

​

,

2

π

​

)

y

1

​

=sinx+cosx

y

2

​

=∣cosx−sinx∣

⇒Area

=∫

0

π/2

​

(y

1

​

−y

2

​

)dx

= ∫

0

π/4

​

 ((sinx+cosx)−(cosx−sinx))dx +∫

π/4

π/2

​

((sinx+cosx)−(sinx−cosx))dx

=4−2

2

​