To compute the triple integral ∭ Ω x 2 d x d y d z \iiint_{\Omega} x^2 \, dx\,dy\,dz ∭ Ω x 2 d x d y d z Ω \Omega Ω z = 3 − x 2 − y 2 z = 3 - x^2 - y^2 z = 3 − x 2 − y 2 z = 0 z = 0 z = 0
Step 1: Convert to Cylindrical Coordinates
In cylindrical coordinates:
x = r cos θ x = r \cos \theta x = r cos θ y = r sin θ y = r \sin \theta y = r sin θ x 2 = r 2 cos 2 θ x^2 = r^2 \cos^2 \theta x 2 = r 2 cos 2 θ The Jacobian (volume element) is r d z d r d θ r\,dz\,dr\,d\theta r d z d r d θ
Step 2: Determine the Limits of Integration
r r r 0 0 0 3 \sqrt{3} 3 x 2 + y 2 = r 2 ≤ 3 x^2 + y^2 = r^2 \leq 3 x 2 + y 2 = r 2 ≤ 3 θ \theta θ 0 0 0 2 π 2\pi 2 π z z z 0 0 0 3 − r 2 3 - r^2 3 − r 2
Step 3: Set Up the Integral
∭ Ω x 2 d x d y d z = ∫ 0 2 π ∫ 0 3 ∫ 0 3 − r 2 r 2 cos 2 θ ⋅ r d z d r d θ = ∫ 0 2 π ∫ 0 3 r 3 cos 2 θ ( 3 − r 2 ) d r d θ \iiint_{\Omega} x^2 \, dx\,dy\,dz = \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{0}^{3 - r^2} r^2 \cos^2 \theta \cdot r \, dz\,dr\,d\theta = \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} r^3 \cos^2 \theta (3 - r^2) \, dr\,d\theta ∭ Ω x 2 d x d y d z = ∫ 0 2 π ∫ 0 3 ∫ 0 3 − r 2 r 2 cos 2 θ ⋅ r d z d r d θ = ∫ 0 2 π ∫ 0 3 r 3 cos 2 θ ( 3 − r 2 ) d r d θ Step 4: Integrate with Respect to z z z
Since the integrand does not depend on z z z z z z
∫ 0 3 − r 2 d z = 3 − r 2 \int_{0}^{3 - r^2} dz = 3 - r^2 ∫ 0 3 − r 2 d z = 3 − r 2 Step 5: Simplify and Integrate with Respect to r r r
Expand the integrand:
r 3 cos 2 θ ( 3 − r 2 ) = 3 r 3 cos 2 θ − r 5 cos 2 θ r^3 \cos^2 \theta (3 - r^2) = 3r^3 \cos^2 \theta - r^5 \cos^2 \theta r 3 cos 2 θ ( 3 − r 2 ) = 3 r 3 cos 2 θ − r 5 cos 2 θ Integrate with respect to r r r
∫ 0 3 ( 3 r 3 − r 5 ) d r = [ 3 r 4 4 − r 6 6 ] 0 3 = 27 4 − 9 2 = 9 4 \int_{0}^{\sqrt{3}} \left(3r^3 - r^5\right) dr = \left[ \frac{3r^4}{4} - \frac{r^6}{6} \right]_0^{\sqrt{3}} = \frac{27}{4} - \frac{9}{2} = \frac{9}{4} ∫ 0 3 ( 3 r 3 − r 5 ) d r = [ 4 3 r 4 − 6 r 6 ] 0 3 = 4 27 − 2 9 = 4 9 Step 6: Integrate with Respect to θ \theta θ
Recall that cos 2 θ = 1 + cos 2 θ 2 \cos^2 \theta = \frac{1 + \cos 2\theta}{2} cos 2 θ = 2 1 + c o s 2 θ
∫ 0 2 π cos 2 θ d θ = 1 2 ∫ 0 2 π ( 1 + cos 2 θ ) d θ = 1 2 ( 2 π + 0 ) = π \int_{0}^{2\pi} \cos^2 \theta \, d\theta = \frac{1}{2} \int_{0}^{2\pi} (1 + \cos 2\theta) \, d\theta = \frac{1}{2} (2\pi + 0) = \pi ∫ 0 2 π cos 2 θ d θ = 2 1 ∫ 0 2 π ( 1 + cos 2 θ ) d θ = 2 1 ( 2 π + 0 ) = π Step 7: Compute the Final Result
Multiply the results of the r r r θ \theta θ
9 4 ⋅ π = 9 π 4 \frac{9}{4} \cdot \pi = \frac{9\pi}{4} 4 9 ⋅ π = 4 9 π Answer:
9 4 π \displaystyle \frac{9}{4}\, \pi 4 9 π
