# Find the volume of the solid enclosed by the surface $x=a\cos^3 u, y=a\sin^3 u, z=v, (0 \le u \le \pi, v \ge 0)$ and the planes $z=0$ and $x+y+z=a$

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.

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.