Calculate this triple integral over D: ∫∫∫10+y2−x2dxdydz over the region bounded D: D={(x,y,z)∣x2+z2≤y2+10, x2+y2≤9}

First of all we see from x2+y2=9 that y lives in Jy=[−3,3] Now we can compute using Fubini: ∫∫∫D10+y2−x2dxdydz =∫y∈[−3,3]dy∫x∈Jx(y)dx∫z∈Jz(x,y)10+y2−x2 =∫y∈[−3,3]dy∫x∈Jx(y)dx⋅210+y2−x2⋅10+y2−x2 =2∫y∈[−3,3]dy∫x∈Jx(y)dx⋅(10+y2−x2) =2∫y∈[−3,3]dy[10x+xy2−13x3]x=−9−y2x=+9+y2 =4∫y∈[−3,3]dy9−y2(10+y2−13(9−y2)) =4∫−339−y2(7+43y2)dy Substitution y=3sin⁡t =4∫−π2π23cos⁡t(7+439sin2t)3cos⁡tdt

"Calculate this triple integral over D: ∫∫∫10+y2−x2dxdydz over..." was answered by our community

lenlifiauw2

Answered question

2021-11-13

Calculate this triple integral over D:

\int \int \int \sqrt{10 + y^{2} - x^{2}} d x d y d z

over the region bounded D:

D = {(x, y, z) ∣ x^{2} + z^{2} \leq y^{2} + 10, x^{2} + y^{2} \leq 9}

Answer & Explanation

Uersfeldte

Beginner2021-11-14Added 20 answers

First of all we see from

x^{2} + y^{2} = 9

that y lives in

J_{y} = [- 3, 3]

Now we can compute using Fubini:

\int \int \int_{D} \sqrt{10 + y^{2} - x^{2}} d x d y d z

= \int_{y \in [- 3, 3]} d y \int_{x \in J_{x} (y)} d x \int_{z \in J_{z} (x, y)} \sqrt{10 + y^{2} - x^{2}}

= \int_{y \in [- 3, 3]} d y \int_{x \in J_{x} (y)} d x \cdot 2 \sqrt{10 + y^{2} - x^{2}} \cdot \sqrt{10 + y^{2} - x^{2}}

= 2 \int_{y \in [- 3, 3]} d y \int_{x \in J_{x} (y)} d x \cdot (10 + y^{2} - x^{2})

= 2 \int_{y \in [- 3, 3]} d y {[10 x + x y^{2} - \frac{1}{3} x^{3}]}_{x = - \sqrt{9 - y^{2}}}^{x = + \sqrt{9 + y^{2}}}

= 4 \int_{y \in [- 3, 3]} d y \sqrt{9 - y^{2}} (10 + y^{2} - \frac{1}{3} (9 - y^{2}))

= 4 \int_{- 3}^{3} \sqrt{9 - y^{2}} (7 + \frac{4}{3} y^{2}) d y

Substitution

y = 3 \sin t

= 4 \int_{- \frac{π}{2}}^{\frac{π}{2}} 3 \cos t (7 + \frac{4}{3} 9 \sin^{2} t) 3 \cos t d t

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