Chapter 13: Q. 22 (page 1082)

Using polar coordinates to evaluate iterated integrals: Evaluate the given iterated integrals by converting them to polar coordinates. Include a sketch of the region.
$\int_{0}^{4} \int_{0}^{\sqrt{4 y - y^{2}}} \frac{1}{\sqrt{x^{2} + y^{2}}} d x d y$

Short Answer

Expert verified

$\int_{0}^{4} \int_{0}^{\sqrt{4 y - y^{2}}} \frac{1}{\sqrt{x^{2} + y^{2}}} d x d y = π$

Step by step solution

Draw the region

From the limits of integration, the region is shown below,

Convert into polar form

By using the below substitution,

$x = r \cos θ y = r \sin θ x^{2} + y^{2} = r^{2} d x d y = r d r d θ$

The equivalent polar integral of the given integral is,

$\int_{0}^{4} \int_{0}^{\sqrt{4 y - y^{2}}} \frac{1}{\sqrt{x^{2} + y^{2}}} d x d y \Rightarrow \int_{0}^{π / 2} \int_{2}^{4} d r d θ$

Calculate integral

$I = \int_{0}^{π / 2} \int_{2}^{4} d r d θ I = {[r]}_{2}^{4} \int_{0}^{π / 2} d θ I = 2 {[θ]}_{0}^{π / 2} I = 2 \times \frac{π}{2} I = π$

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Using polar coordinates to evaluate iterated integrals: Evaluate the given iterated integrals by converting them to polar coordinates. Include a sketch of the region.
$\int_{0}^{4} \int_{0}^{\sqrt{4 y - y^{2}}} \frac{1}{\sqrt{x^{2} + y^{2}}} d x d y$

Short Answer

Step by step solution

Draw the region

Convert into polar form

Calculate integral

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Using polar coordinates to evaluate iterated integrals: Evaluate the given iterated integrals by converting them to polar coordinates. Include a sketch of the region.∫04∫04y-y21x2+y2dxdy

Short Answer

Step by step solution

Draw the region

Convert into polar form

Calculate integral

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Geometry

Decision Maths

Statistics

Discrete Mathematics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Using polar coordinates to evaluate iterated integrals: Evaluate the given iterated integrals by converting them to polar coordinates. Include a sketch of the region.
$\int_{0}^{4} \int_{0}^{\sqrt{4 y - y^{2}}} \frac{1}{\sqrt{x^{2} + y^{2}}} d x d y$