Question

Use cylindrical coordinates.Evaluatex2 dV,iiintegral.gifEwhere E is the solid that lies within the cylinderx2 + y2 = 4,above the planez = 0,and below the conez2 = 16x2 + 16y2.

Answer 1

The value of the integral ∫x²dV using cylindrical coordinates is 512π / 3 .

In this case, the solid is positioned between the cylinder x² + y² = 4 and will be evaluated by using the cone z²= 16x² + 16y² .

Cylindric coordinates will be used to calculate this integral  $\int\limits_e {x^2} \, dx$.

The distance from the origin to the point (x, y, 0) and the angle the point (x, y, 0) makes with the x-axis, respectively, are provided by the formulas.

x = r cos θ

y= r sin θ

With these coordinates, we shall characterize the solid E.

The boundary is then x² + y² = r²

The boundary also becomes r² = 4 or ( 0 ≤ θ ≤ r² )

Again z = 16r²

The original integral must now be translated into terms of these new coordinates. It is necessary to multiply the function by the Jacobian of the change in coordinates in order to ensure that the result remains the same. The value for the cylindric coordinates is r. Then

$\int_ex^2 dx = \int\limits^{2\pi}_0\int\limits^2_0\int\limits^{16r^2}_0 r(r^2cos^2\theta)dzdrd\theta$

$=\int _0^{2\pi }\int _0^216r^5\cos ^2\theta drd\theta$

$=\int _0^{2\pi }\frac{512}{3}\cos ^2\theta d\theta$

= 512/3 π

Hence the volume of the solid that is enclosed by the integral is given by 512/3 π .

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