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