The value of -x-01-y-0x2-z-0yy-2zdz dy dx is

I=∫_x=0^1∫_y=0^x^2∫_z=0^y(y+2z)dx dy dx nbsp; =∫_0^1∫_0^x^2(yz+z^2)^y_0dy dx =∫_0^1∫_0^x^2(2y^2)dy dx =∫_0^1 (2y^33)^x^2_0 dx =∫_0^12x^63dx =[2x^721 ]^1_ ...

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Engineering Mathematics

Double integrals and Triple Integrals

The value of ...

Question

The value of $\int_{x = 0}^{1} \int_{y = 0}^{x^{2}} \int_{z = 0}^{y} (y + 2 z) d z d y d x$ is

\frac{1}{53}

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\frac{2}{21}

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\frac{1}{6}

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\frac{5}{3}

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Solution

The correct option is B $\frac{2}{21}$

$I = \int_{x = 0}^{1} \int_{y = 0}^{x^{2}} \int_{z = 0}^{y} (y + 2 z) d x d y d x$

= $\int_{0}^{1} \int_{0}^{x^{2}} (y z + z^{2})_{0}^{y} d y d x$

= $\int_{0}^{1} \int_{0}^{x^{2}} (2 y^{2}) d y d x$

= $\int_{0}^{1} {(\frac{2 y^{3}}{3})}_{0}^{x^{2}} d x$

= $\int_{0}^{1} \frac{2 x^{6}}{3} d x$

= ${[\frac{2 x^{7}}{21}]}_{0}^{1}$

= $\frac{2}{21}$

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The value of ∫1x=0∫x2y=0∫yz=0(y+2z)dz dy dx is

The correct option is B 221 I=∫1x=0∫x2y=0∫yz=0(y+2z)dxdydx =∫10∫x20(yz+z2)y0dydx =∫10∫x20(2y2)dydx =∫10(2y33)x20dx =∫102x63dx =[2x721]10 =221

The value of $\int_{x = 0}^{1} \int_{y = 0}^{x^{2}} \int_{z = 0}^{y} (y + 2 z) d z d y d x$ is