Chapter 13: Q. 37 (page 1015)

In Exercises 35–40, find the volume of the solid bounded above by the given function over the specified region $Ω$ . $f (x, y) = \sqrt{4 - x^{2} - y^{2}}$
Region:

Short Answer

Expert verified

Volume bounded by given function is $4 π$ .

Step by step solution

Step 1. Given Information

A function, $f (x, y) = \sqrt{4 - x^{2} - y^{2}}$

Region:

Step 2. Calculating the volume of the solid bounded above by the given function and region

The double integral is given by,

$\int_{- 2}^{2} \int_{0}^{\sqrt{4 - x^{2}}} \sqrt{4 - x^{2} - y^{2}} d y d x + \int_{- 2}^{0} \int_{- \sqrt{4 - x^{2}}}^{0} \sqrt{4 - x^{2} - y^{2}} d y d x$

Due to symmetry it can also be written as,

$3 \int_{0}^{2} \int_{0}^{\sqrt{4 - x^{2}}} \sqrt{4 - x^{2} - y^{2}} d y d x$

Converting to polar coordinates, the integration becomes,

$3 \int_{0}^{\frac{π}{2}} \int_{0}^{2} (\sqrt{4 - r^{2}}) r d r d θ$

Put, $4 - r^{2} = t, - 2 r d r = d t,$

We get, $3 \int_{0}^{\frac{π}{2}} \int_{4}^{0} (- \frac{\sqrt{t}}{2}) d t d θ$

$= - \frac{3}{2} \int_{0}^{\frac{π}{2}} \frac{2}{3} {(t^{\frac{3}{2}})}_{4}^{0} d θ = - \int_{0}^{\frac{π}{2}} {(0 - 4^{\frac{3}{2}})}_{4}^{0} d θ = - 3 \int_{0}^{\frac{π}{2}} \frac{2}{3} (0 - 4^{\frac{3}{2}}) d θ = 8 \int_{0}^{\frac{π}{2}} 1 d θ = 8 θ_{0}^{\frac{π}{2}} = 8 \times \frac{π}{2} = 4 π$

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In Exercises 35–40, find the volume of the solid bounded above by the given function over the specified region $Ω$ . $f (x, y) = \sqrt{4 - x^{2} - y^{2}}$
Region:

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculating the volume of the solid bounded above by the given function and region

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In Exercises 35–40, find the volume of the solid bounded above by the given function over the specified regionΩ. f(x,y)=4-x2-y2Region:

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculating the volume of the solid bounded above by the given function and region

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Calculus

Discrete Mathematics

Pure Maths

Mechanics Maths

Statistics

Study anywhere. Anytime. Across all devices.

In Exercises 35–40, find the volume of the solid bounded above by the given function over the specified region $Ω$ . $f (x, y) = \sqrt{4 - x^{2} - y^{2}}$
Region: