SOLVED:39-41 Evaluate the integral by changing to spherical coordinates. ∫0^1 ∫0^√(1-x^3) ∫√(x^2+y^2)^√(2-x^2-y^2) x y d z d y d x

Evaluate the integral by changing to spherical coordinates – Problem 15.8.43 Cengage Calculus
Evaluate the integral by changing to spherical coordinates – Problem 15.8.43 Cengage Calculus

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Chapter 15, Problem 39

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$39-41$ Evaluate the integral by changing to spherical coordinates.$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{3}}} \int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2}-y^{2}}} x y d z d y d x$

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08:56

$37-39$ Evaluate the integral by changing to spherical coordinates.
$$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2…

07:03

$39-41$ Evaluate the integral by changing to spherical coordinates.
$\int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \int_{2-\sqrt{4-x^{2}-y^{…

03:30

$39-41$ Evaluate the integral by changing to spherical coordinates.
$\int_{-a}^{a} \int_{-\sqrt{a^{2}-y^{2}}}^{\sqrt{a^{2}-y^{2}}} \int_{-\sqrt{a^{2…

06:05

$37-39$ Evaluate the integral by changing to spherical coordinates.
$$\int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \int_{2-\sqrt{4-x^{2}-y^…

05:34

$37-39$ Evaluate the integral by changing to spherical coordinates.
$$\int_{-a}^{a} \int_{-\sqrt{a^{2}-y^{2}}}^{\sqrt{a^{2}-y^{2}}} \int_{-\sqrt{a^{…

08:40

Evaluate the integral by changing to spherical coordinates.
$$
\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{\sqrt{x^{3}+y^{2}}}^{\sqrt{2-x^{2}-y^{2…

03:03

Evaluate the integral by changing to spherical coordinates.
$\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2}-y^{2}}} …

07:34

Evaluate the integral by changing to spherical coordinates.
$ \displaystyle \int_0^1 \int_0^{\sqrt{1 – x^2}} \int_{\sqrt{x^2 + y^2}}^{\sqrt{2 – x^…

Transcript

were given an integral and were asked to evaluate this integral by changing to spherical coordinates. This is the integral from 01 integral from zero to the square root of one minus X cubed, integral from the square of X squared plus y squared to the square, two to minus X squared, minus y squared of x times y B C D Y d efs Well, first of all region integration mean is above the cone. Z equals the square root of X squared plus y squared and below the sphere Z equals square root of two minus X squared, minus y squared. Or, in other words, X squared. Plus y squared plus C squared equals two. And this lies in the first document. Since you see that X lies between zero and one and wildlife between zero and the positive square of one minus X squared so it cuts in the first doctrines. It follows that data live between zero in a pi over two, he cone, she pulls the square x squared plus y squared. This has the equation. Fi equals pi over four, and so we have that fine lines between zero and pi over four since we’re looking at the region above the cone and ro is going toe. Lie between zero and the radius of the sphere, which is the square root of two. So the integral becomes integral from bye equals zero pi over four. Integral from 30 equals zero to pi over two, integral from zero equals zero to root. Two our function, which was X y. But now we used spherical coordinates. This is row sine phi cosine theta times row signed by sign data times…

You are watching: SOLVED:39-41 Evaluate the integral by changing to spherical coordinates. ∫0^1 ∫0^√(1-x^3) ∫√(x^2+y^2)^√(2-x^2-y^2) x y d z d y d x. Info created by THVinhTuy selection and synthesis along with other related topics.

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