Chapter 13: Q. 59 (page 1028)

Sketch the region of integration for each of integrals in Exercises $57 - 60$ , and then evaluate the integral by converting to polar coordinates.
$\int_{0}^{4} \int_{0}^{\sqrt{16 - x^{2}}} e^{x^{2}} e^{y^{2}} d y d x$

Short Answer

Expert verified

The value of the integral is $\int_{0}^{4} \int_{0}^{\sqrt{16 - x^{2}}} e^{x^{2}} e^{y^{2}} d y d x = \frac{π}{4} (e^{16} - 1)$

Step by step solution

Given information

The integral is $I = \int_{0}^{4} \int_{0}^{\sqrt{16 - x^{2}}} e^{x^{2}} e^{y^{2}} d y d x$

Here, $x = 0$ and $x = 4, y = 0$ and $y = \sqrt{16 - x^{2}}$

Calculation

The region of integration R is shown in the figure

$r^{2} \sin^{2} θ + r^{2} \cos^{2} θ = 16 r^{2} = 16 \Rightarrow r = 4$

Substitute $x = r \cos θ$ in the lower limit of $x$ .

$r \cos θ = 0 \Rightarrow r = 0, θ = \frac{π}{2}$

Thus, the limits of $r$ are $r = 0$ and $r = 4$ and that of $θ$ are $0$ and $\frac{π}{2} .$

$d x d y = r d r d θ$

Therefore,

$I = \int_{0}^{4} \int_{0}^{\sqrt{16 - x^{2}}} e^{x^{2}} e^{y^{2}} d y d x = \int_{0}^{x / 2} \int_{0}^{4} e^{r^{2}} r d r d θ I = \int_{0}^{π / 2} [\int_{0}^{4} r e^{r^{2}} d r] d θ$

Integrate with respect to $r$ first

Put $r^{2} = t$

$2 r d r = d t \Rightarrow r d r = \frac{d t}{2} I = \int_{0}^{π / 2} [\int_{0}^{16} e^{t} \frac{d t}{2}] d θ I = \frac{1}{2} \int_{0}^{π / 2} {[e^{t}]}_{0}^{16} d θ I = \frac{1}{2} \int_{0}^{π / 2} [e^{16} - e^{0}] d θ I = \frac{1}{2} \int_{0}^{π / 2} [e^{16} - 1] d θ I = \frac{1}{2} (e^{16} - 1) [θ]_{0}^{π / 2} I = \frac{1}{2} (e^{16} - 1) [\frac{π}{2}] I = \frac{π}{4} (e^{16} - 1)$

Thus, the value of the integral is

$\int_{0}^{4} \int_{0}^{\sqrt{16 - x^{2}}} e^{x^{2}} e^{y^{2}} d y d x = \frac{π}{4} (e^{16} - 1)$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Sketch the region of integration for each of integrals in Exercises $57 - 60$ , and then evaluate the integral by converting to polar coordinates.
$\int_{0}^{4} \int_{0}^{\sqrt{16 - x^{2}}} e^{x^{2}} e^{y^{2}} d y d x$

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Pure Maths

Geometry

Statistics

Decision Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Sketch the region of integration for each of integrals in Exercises 57–60, and then evaluate the integral by converting to polar coordinates.∫04∫016-x2ex2ey2dydx

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Pure Maths

Geometry

Statistics

Decision Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Sketch the region of integration for each of integrals in Exercises $57 - 60$ , and then evaluate the integral by converting to polar coordinates.
$\int_{0}^{4} \int_{0}^{\sqrt{16 - x^{2}}} e^{x^{2}} e^{y^{2}} d y d x$