Chapter 5: Q15P (page 268)

Express the integral $I = \int_{0}^{1} d x \int_{0}^{\sqrt{1 - x^{2}}} e^{- x^{2} - y^{2}} d y$ as an integral in polar coordinates $(r, θ)$ and so evaluate it.

Short Answer

Expert verified

The value of Integral is $I = \frac{ττ}{4} (1 - e^{- 1})$

Step by step solution

Given Integral

The given integral is, $I = \int_{0}^{1} dx \int_{0}^{\sqrt{1 - x^{2}}} e^{- x^{2} - y^{2}} dy$

Concept of converting integral to polar coordinates

Use $x = r c o s$ , $r = \sin θ$ and $d A = r d r d θ$ to convert an integral in rectangular coordinates to an integral in polar coordinates.

Evaluate the integral

It is required to express the integral as an integral in polar coordinates $(r, θ)$ and evaluate it.

The area of integration is the quarter-circle in the first quadrant, of unit radius.

Thus, the integral become:

$I = \int_{0}^{1} dx \int_{0}^{\sqrt{1 - x^{2}}} e^{- x^{2} - y^{2}} dy I = \int_{r = 0}^{1} \int_{θ = 0}^{τ τ / 2} e^{- r} rdrdθ I = \frac{ττ}{2} \int_{0}^{1} e^{- r^{2}} d (- r^{2}) (\frac{1}{2}) I = {|- \frac{ττ}{4} e^{- r^{2}}|}_{0}^{1} I = \frac{ττ}{4} (1 - e^{- 1})$

The value of Integral is $I = \frac{ττ}{4} (1 - e^{- 1})$ .

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Express the integral $I = \int_{0}^{1} d x \int_{0}^{\sqrt{1 - x^{2}}} e^{- x^{2} - y^{2}} d y$ as an integral in polar coordinates $(r, θ)$ and so evaluate it.

Short Answer

Step by step solution

Given Integral

Concept of converting integral to polar coordinates

Evaluate the integral

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Express the integral I=∫01dx∫01-x2e-x2-y2dyas an integral in polar coordinates (r,θ) and so evaluate it.

Short Answer

Step by step solution

Given Integral

Concept of converting integral to polar coordinates

Evaluate the integral

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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Study anywhere. Anytime. Across all devices.

Express the integral $I = \int_{0}^{1} d x \int_{0}^{\sqrt{1 - x^{2}}} e^{- x^{2} - y^{2}} d y$ as an integral in polar coordinates $(r, θ)$ and so evaluate it.