Chapter 9: Q. 37 (page 775)

Areas of regions bounded by polar functions: Find the areas of the following regions. The area bounded by one loop of $r^{2} = - \sin 2 θ$

Short Answer

Expert verified

The required area is $\frac{3}{4} units$

Step by step solution

Given information

$r^{2} = - \sin 2 θ$

Area

The graph of the given function $r^{2} = - \sin 2 θ$ is,

The area bounded by one loop is,

$A = \frac{1}{2} \int_{α}^{β} r^{2} d θ A = \frac{1}{2} \int_{0}^{\frac{π}{2}} - \sin 2 θ d θ A = \frac{1}{2} {(\frac{\cos 2 θ}{2})}_{0}^{\frac{π}{2}} A = \frac{1}{2} (\frac{\cos 2 (\frac{π}{2})}{2} - \cos 2 (0)) A = \frac{1}{2} (\frac{- 1}{2} - 1) A = \frac{3}{4} units$

The required area is $A = \frac{3}{4} units$

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Areas of regions bounded by polar functions: Find the areas of the following regions. The area bounded by one loop of $r^{2} = - \sin 2 θ$

Short Answer

Step by step solution

Given information

Area

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Areas of regions bounded by polar functions: Find the areas of the following regions. The area bounded by one loop of r2=−sin2θ

Short Answer

Step by step solution

Given information

Area

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Geometry

Calculus

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Study anywhere. Anytime. Across all devices.

Areas of regions bounded by polar functions: Find the areas of the following regions. The area bounded by one loop of $r^{2} = - \sin 2 θ$