The figure shows the region of integration for the integral. r6 36-x (6-x L Fx, y, 2) dy dz dx Rewrite this integral as an equivalent iterated integral in the five other orders. (Assume y(x) = 6 - x and z(x) = 36 – x2.)
The figure shows the region of integration for the integral. r6 36-x (6-x L Fx, y, 2) dy dz dx Rewrite this integral as an equivalent iterated integral in the five other orders. (Assume y(x) = 6 - x and z(x) = 36 – x2.)
Calculus: Early Transcendentals
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![The figure shows the region of integration for the integral:
\[
\int_0^6 \int_0^{36-x^2} \int_0^{6-x} f(x, y, z) \, dy \, dz \, dx
\]
Rewrite this integral as an equivalent iterated integral in the five other orders. (Assume \( y(x) = 6 - x \) and \( z(x) = 36 - x^2 \).)
**Diagram Explanation:**
The image includes a diagram of a three-dimensional region bounded by the curves \( z(x) \) and \( y(x) \), illustrating the volume over which the function \( f(x, y, z) \) is integrated.
**Equivalent Iterated Integrals:**
1. \[
\int_0^6 \int_0^{6-x} \int_0^{36-x^2} f(x, y, z) \, dy \, dx \, dz
\]
2. \[
\int_0^6 \int_0^{36-x^2} \int_0^{6-x} f(x, y, z) \, dz \, dx \, dy
\]
3. \[
\int_0^6 \int_0^{6-x} \int_0^{36-x^2} f(x, y, z) \, dz \, dy \, dx
\]
4. \[
\int_0^6 \int_0^{36-x^2} \int_0^{6-x} f(x, y, z) \, dx \, dy \, dz
\]
5. \[
\int_0^6 \int_0^{\sqrt{36-z}} \int_0^{6-\sqrt{36-z}} f(x, y, z) \, dx \, dy \, dz
\]
6. \[
\int_0^{36} \int_0^{\sqrt{36-y^2}} \int_0^{6-y} f(x, y, z) \, dx \, dz \, dy
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1c7104a7-391d-43e8-bb65-99aa718e4fa3%2F5764fc6b-c068-4b85-8d11-6e03427cad36%2Fcvua6ee_processed.png&w=3840&q=75)
Transcribed Image Text:The figure shows the region of integration for the integral:
\[
\int_0^6 \int_0^{36-x^2} \int_0^{6-x} f(x, y, z) \, dy \, dz \, dx
\]
Rewrite this integral as an equivalent iterated integral in the five other orders. (Assume \( y(x) = 6 - x \) and \( z(x) = 36 - x^2 \).)
**Diagram Explanation:**
The image includes a diagram of a three-dimensional region bounded by the curves \( z(x) \) and \( y(x) \), illustrating the volume over which the function \( f(x, y, z) \) is integrated.
**Equivalent Iterated Integrals:**
1. \[
\int_0^6 \int_0^{6-x} \int_0^{36-x^2} f(x, y, z) \, dy \, dx \, dz
\]
2. \[
\int_0^6 \int_0^{36-x^2} \int_0^{6-x} f(x, y, z) \, dz \, dx \, dy
\]
3. \[
\int_0^6 \int_0^{6-x} \int_0^{36-x^2} f(x, y, z) \, dz \, dy \, dx
\]
4. \[
\int_0^6 \int_0^{36-x^2} \int_0^{6-x} f(x, y, z) \, dx \, dy \, dz
\]
5. \[
\int_0^6 \int_0^{\sqrt{36-z}} \int_0^{6-\sqrt{36-z}} f(x, y, z) \, dx \, dy \, dz
\]
6. \[
\int_0^{36} \int_0^{\sqrt{36-y^2}} \int_0^{6-y} f(x, y, z) \, dx \, dz \, dy
\]
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