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Chapter 13: Problem 5

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} y d y d x $$

Short Answer

Expert verified
The region of integration is a semi-circle in the first quadrant. The value of the double integral is \(\frac{1}{3}\).

Step by step solution

01

Understand the region of integration

The region of integration is given by the bounds of the integral: \(y\) is bounded between 0 and \(\sqrt{1-x^2}\) and \(x\) is between 0 and 1. This represents a semi-circle in the first quadrant.
02

Evaluate the inner integral

The inner integral can be evaluated as: \[ \int_{0}^{\sqrt{1-x^{2}}} y d y = [\frac{y^2}{2}]_{0}^{\sqrt{1-x^{2}}} = \frac{1}{2} (1 - x^2) \] After substituting the limits of \(y\), we get a function in terms of \(x\).
03

Evaluate the outer integral

Now, integrate the result of inner integral with respect to \(x\): \[ \int_{0}^{1} \frac{1}{2}(1-x^{2}) dx = [\frac{x}{2} - \frac{x^3}{6}]_{0}^{1} = \frac{1}{3} \]

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Most popular questions from this chapter

Evaluate the double integral. $$ \int_{0}^{2} \int_{3 y^{2}-6 y}^{2 y-y^{2}} 3 y d x d y $$

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{0}^{2} \int_{x / 2}^{1} d y d x $$

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Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{0}^{1} \int_{0}^{2} d y d x $$
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