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Calculus Level 5

If the integral ∫ 0 1 ∫ 0 1 − x 2 ∫ x 2 + y 2 1 1 x 2 + y 2 + z 2 dz dy dx \displaystyle\int_0^{1}\displaystyle\int_0^{\sqrt{1-x^2}}\displaystyle\int_{\sqrt{x^2+y^2}}^{1} \dfrac{1}{\sqrt{x^2+y^2+z^2}} \text{dz dy dx} can be represented as π ( a − b ) c \dfrac{\pi(\sqrt{\text{a}} - \text{b})}{\text{c}} . Find the value of a+b+c \text{a+b+c} .

a a is a square free integer, b b and c c are coprime integers.


The answer is 7.

Parth Lohomi by Parth Lohomi , 20, India

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1 solution

Mark Hennings
Apr 7, 2018

Converting the integral over x x and y y to polar coordinates, the integral is X = ∫ 0 1 d x ∫ 0 1 − x 2 d y ∫ x 2 + y 2 1 d z x 2 + y 2 + z 2 = ∫ 0 1 d r ∫ 0 1 2 π d θ ∫ r 1 r d z r 2 + z 2 = 1 2 π ∫ 0 1 d r ∫ r 1 r d z r 2 + z 2 = 1 2 π ∫ 0 1 d z ∫ 0 r r d r r 2 + z 2 = 1 2 π ∫ 0 1 d z [ r 2 + z 2 ] r = 0 z = 1 2 π ( 2 − 1 ) ∫ 0 1 z d z = 1 4 π ( 2 − 1 ) \begin{aligned} X & = \; \int_0^1\,dx \int_0^{\sqrt{1-x^2}}\,dy \int_{\sqrt{x^2+y^2}}^1 \frac{dz}{\sqrt{x^2+y^2+z^2}} \; = \; \int_0^1\,dr \int_0^{\frac12\pi}\,d\theta \int_r^1 \frac{r\,dz}{\sqrt{r^2 + z^2}} \\ & = \; \tfrac12\pi \int_0^1\,dr \int_r^1\frac{r\,dz}{\sqrt{r^2 + z^2}} \; = \; \tfrac12\pi\int_0^1\,dz \int_0^r \frac{r\,dr}{\sqrt{r^2 + z^2}} \\ & = \; \tfrac12\pi \int_0^1\,dz \Big[\sqrt{r^2 + z^2}\Big]_{r=0}^z \; = \; \tfrac12\pi(\sqrt{2}-1)\int_0^1 z\,dz \\ & = \; \tfrac14\pi(\sqrt{2}-1) \end{aligned} which makes the answer 2 + 1 + 4 = 7 2 + 1 + 4 = \boxed{7} .

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You can also convert it into spherical coordinates.

Parth Lohomi - 3 years, 2 months ago

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True, but the integrand is not that neat over a cone.

Mark Hennings - 3 years, 2 months ago

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Yes that was the main reason of posting the question,I never thought of polar substitution.

Parth Lohomi - 3 years, 2 months ago

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@Parth Lohomi – Well, I suppose you can use polar coordinates, for 0 ≤ ϕ ≤ 1 2 π 0\le\phi\le\tfrac12\pi , 0 ≤ r ≤ sec ⁡ θ 0\le r\le \sec\theta , 0 ≤ θ ≤ 1 4 π 0\le\theta\le\tfrac14\pi . The r r integral is easy, ∫ r − 1 r 2 d r = 1 2 r 2 \int r^{-1}\, r^2\, dr = \tfrac12r^2 leading to a θ \theta integral ∫ 0 1 4 π 1 2 sec ⁡ θ tan ⁡ θ d θ = 1 2 ( 2 − 1 ) \int_0^{\frac14\pi}\tfrac12\sec\theta \tan\theta\,d\theta = \tfrac12(\sqrt{2}-1) with the ϕ \phi integral being trivial.

Mark Hennings - 3 years, 2 months ago

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@Mark Hennings – I meant that I never thought of using polar coordinates, not that I don't think we can use it!

Parth Lohomi - 3 years, 2 months ago

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