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Rewrite the integral \int_0^1 \int_0^{1-x^2}\int_0^{1-x} f(x,y,z) \, dy \,dz \,dx as an...

Question:

Rewrite the integral {eq}\int_0^1 \int_0^{1-x^2}\int_0^{1-x} f(x,y,z) \, dy \,dz \,dx {/eq} as an equivalent integral in five other forms.

Triple Integral:

Triple integral is the function of three dimensional space.Triple integral defined by six different orders. It is used to find the area, volume and mass of the solid.Therefore: Triple integral can be written as:

{eq}I= \int \int \int f(x,y,z) d x d y dz. {/eq}

Answer and Explanation:

Given:

a) {eq}\int_0^1 \int_0^{1-x^2}\int_0^{1-x} f(x,y,z) \, dy \,dz \,dx {/eq}

b) Now: we write the limits of triple integral d x d y dz such that:

{eq}0 \leq x \leq 1-y, \enspace 0 \leq y \leq \sqrt{1-z} \enspace 0 \leq z \leq 1 {/eq}

Hence:

{eq}Volume = \int_{0}^{1}\int_{0}^{\sqrt{1-z}}\int_{0}^{1-y} dx dy dz {/eq}

c) Now: we write the limits of dx dz dy

{eq}0 \leq y \leq 1 , \enspace 0 \leq x \leq \sqrt{1-z}, \enspace 0 \leq z \leq 1-y {/eq}

Hence:

{eq}Volume = \int_{0}^{1}\int_{0}^{1-y }\int_{0}^{1-y} dx dy dz {/eq}

d) Now write dz dx dy

limits: {eq}0 \leq y \leq 1 , \enspace 0 \leq z \leq 1-x^2, \enspace 0 \leq x \leq 1-y {/eq}

Hence:

{eq}Volume = \int_{0}^{1}\int_{0}^{1-y }\int_{0}^{1-x^2} dz dx dy {/eq}

e) Now: write dz dy dx

limits: {eq}0 \leq y \leq \sqrt{1-z} , \enspace 0 \leq z \leq 1-x^2 , \enspace 0 \leq x \leq 1 {/eq}

Hence:

{eq}Volume = \int_{0}^{1}\int_{0}^{\sqrt{1-z} }\int_{0}^{1-x^2} dz d y d x {/eq}

f) Now write: dy dx dz

Limits: {eq}0 \leq z \leq 1 , \enspace 0 \leq y \leq 1-x , \enspace 0 \leq x \leq \sqrt{1-z} {/eq}

Hence:

{eq}Volume = \int_{0}^{1}\int_{0}^{\sqrt{1-z} }\int_{0}^{1-x} d y d x d z {/eq}


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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