Question #101459
x^2/3+y^2/3=a^2/3 , a>0 (astroid) rotate about y-axis and find the volume?
1
Expert's answer
2020-01-20T09:37:40-0500

V=aaπ((a23y23)32)2dy==πaa(a23y23)3dy==πaa(a23a43y23+3a23y43y2)dy==π(a2y3a4335y53+3a2337y7313y3)aa==π((a395a3+97a313a3)(a3+95a397a3+13a3))==π(16105a3+16105a3)==π32105a3V=\int_{-a}^{a}\pi((a^{\frac {2}{3}}-y^{\frac {2}{3}})^\frac {3}{2})^2dy=\\ =\pi \int_{-a}^{a}(a^{\frac {2}{3}}-y^{\frac {2}{3}})^{3}dy=\\ =\pi\int_{-a}^{a}(a^{2}-3a^{\frac{4}{3}}y^{\frac {2}{3}}+3a^\frac {2}{3}y^{\frac{4}{3}}-y^2)dy=\\ =\pi(a^{2}y-3a^{\frac{4}{3}}\frac{3}{5}y^{\frac {5}{3}}+3a^\frac {2}{3}\frac{3}{7}y^{\frac{7}{3}}-\frac{1}{3}y^3)|^{a}_{-a}=\\ =\pi((a^{3}-{\frac{9}{5}}a^{3}+\frac {9}{7}a^{3}-\frac{1}{3}a^3)-\\-(-a^{3}+{\frac{9}{5}}a^{3}-\frac {9}{7}a^{3}+\frac{1}{3}a^3))=\\ =\pi(\frac{16}{105}a^{3}+\frac{16}{105}a^{3})=\\ =\pi\frac{32}{105}a^{3}


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