GATE (TF) Textile 2018 Question Paper Solution | GATE/2018/TF/41

Question 41 (Textile Technology & Fibre Science)

The value of the integral \int_{0}^{4}\int_{0}^{\sqrt{16-x^2}}y\sqrt{x^2+y^2}\:\:\:dydx is __64__.

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Let, x^2+y^2=v
Now, integrate this-
\int_{0}^{4}\int_{0}^{\sqrt{16-x^2}} \frac{\sqrt{v}}{2}dvdx
First solve-
\frac{1}{2}\int_{0}^{\sqrt{16-x^2}} \sqrt{v} dv
\frac{1}{2}\left [\frac{2v^\frac{3}{2}}{3} \right ]_{0}^{\sqrt{16-x^2}}
\frac{1}{3}\left [v^\frac{3}{2} \right ]_{0}^{\sqrt{16-x^2}}
\frac{1}{3} \left [(\sqrt{16-x^2})^\frac{3}{2} \right ]
Now, solve second integration-
\int_{0}^{4} \frac{(16-x^2)^3}{3} dx
\frac{1}{3}\int_{0}^{4} (16-x^2)^3 dx
\frac{1}{3}\int_{0}^{4} (16-x^2)^3 dx
=64 (Ans)

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