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My attempt:

$$\operatorname{vol}(V) = \iint_D dxdy\int_0^{a-x-y}dz = \iint_D(a-x-y)dxdy,$$ where $D$ is the region that represents the given part of an asteroid ($x,y$).

Now I don't know how to find the double integral from above. What I did:

$$ = 2\int_0^a dx \int_0^{f(x)}(a-x-y)dy.$$

This does not give me the correct answer (unless someone would tell me this is the right approach, meaning I made an error in the integrations and/or substitution...).

The correct answer: $a^2(\frac{19}{210}-\pi\frac{3}{32})$.

Thanks.

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The question is in fact to find the volume enclosed by the surface $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$ with x from $-a$ to $a$ and $x + y + z = a$

Hence the volume integral V should be $$\int_{-a}^{a} \int_{0}^{(a^{\frac{2}{3}} - x^{\frac{2}{3}})^{\frac{3}{2}}} (a - x - y)dy dx$$

Using substitution $x = a sin^{3} \theta$ and $\theta = \frac{\pi}{2}$ to $\theta = \frac{-\pi}{2}$

$$V = 3a^{3} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} (sin^{2}\theta cos^{4}\theta - sin^{5}\theta cos^{4}\theta - \frac{1}{2}sin^{2}\theta cos^{7}\theta)d\theta$$

$$= a^3(\frac{3\pi}{8} - \frac{16}{105})$$

Notice that the second term of the integrand is an odd function and its integral is zero.

Hence it looks like the given answer is incorrect since it has $a^2$ and not $a^3$. Also it is negative.

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