Chapter 13: Q. 18 (page 1055)

Complete Example $6$ by evaluating the iterated integral $\int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{x + z + 4} k y d y d z d x .$

Short Answer

Expert verified

The value of the iterated integral $\int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{x + z + 4} k y d y d z d x$ is, $110 k$ .

Step by step solution

Step 1. Given information

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

Step 2. Evaluate the given triple integral.

$\int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{x + z + 4} k y d y d z d x = \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} (\int_{0}^{x + z + 4} k y d y) d z d x = \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} (k {[\frac{y^{2}}{2}]}_{0}^{x + z + 4}) d z d x = \frac{k}{2} \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} {(x + z + 4)}^{2} d z d x$

Now expand the obtained expression.

$\int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{x + z + 4} k y d y d z d x = \frac{k}{2} \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} (x^{2} + z^{2} + 16 + 2 x z + 8 z + 8 x) d z d x$

Step 3. Now integrate the obtained expression with respect to z.

$\int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{x + z + 4} k y d y d z d x = \frac{k}{2} \int_{- 2}^{2} [x^{2} {[z]}_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} + {[\frac{z^{3}}{3}]}_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} + 16 {[z]}_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} + 2 x {[\frac{z^{2}}{2}]}_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} + 8 {[\frac{z^{2}}{2}]}_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} + 8 x {[z]}_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}}] = \frac{k}{2} \int_{- 2}^{2} [x^{2} (2 \sqrt{4 - x^{2}}) + \frac{2}{3} {(\sqrt{4 - x^{2}})}^{3} + 16 (2 \sqrt{4 - x^{2}}) + x (0) + 4 (0) + 8 x (2 \sqrt{4 - x^{2}})] d x = \frac{k}{2} \int_{- 2}^{2} [2 x^{2} (\sqrt{4 - x^{2}}) + \frac{2}{3} {(4 - x^{2})}^{\frac{3}{2}} + 32 (\sqrt{4 - x^{2}}) + 16 x (\sqrt{4 - x^{2}})] d x$

Step 4. Now integrate with respect to x.

$\int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{x + z + 4} k y d y d z d x = \frac{k}{2} [2 \int_{- 2}^{2} x^{2} (\sqrt{4 - x^{2}}) d x + \frac{2}{3} \int_{- 2}^{2} {(4 - x^{2})}^{\frac{3}{2} d x} + 32 \int_{- 2}^{2} (\sqrt{4 - x^{2}}) d x + 16 \int_{- 2}^{2} x (\sqrt{4 - x^{2}}) d x] = \frac{k}{2} [2 π + \frac{2}{3} (6 π) + 32 (2 π) + 0] = \frac{k}{2} [2 π + 4 π + 64 π] = 35 kπ = 110 k$

Therefore, the required value is, $110 k$ .

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Complete Example $6$ by evaluating the iterated integral $\int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{x + z + 4} k y d y d z d x .$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Evaluate the given triple integral.

Step 3. Now integrate the obtained expression with respect to z.

Step 4. Now integrate with respect to x.

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Step by step solution

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Step 2. Evaluate the given triple integral.

Step 3. Now integrate the obtained expression with respect to z.

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One App. One Place for Learning.

Most popular questions from this chapter

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