Studdy AI Logo
Math  /  Calculus

QuestionV=∫02π∫022∫r36−r2rdzdrdθV=\int_{0}^{2 \pi} \int_{0}^{2 \sqrt{2}} \int_{r}^{\sqrt{36-r^{2}}} r d z d r d \theta

Studdy Solution

STEP 1

1. The given integral represents the volume of a region in cylindrical coordinates.
2. The limits of integration suggest a cylindrical region with specific boundaries in zz, rr, and θ\theta.
3. We will evaluate the integral step-by-step, starting from the innermost integral to the outermost.

STEP 2

1. Evaluate the innermost integral with respect to zz.
2. Evaluate the resulting integral with respect to rr.
3. Evaluate the final integral with respect to θ\theta.

STEP 3

Evaluate the innermost integral with respect to zz:
The innermost integral is:
∫r36−r2r dz \int_{r}^{\sqrt{36-r^{2}}} r \, dz
Since rr is treated as a constant with respect to zz, the integral becomes:
r∫r36−r21 dz r \int_{r}^{\sqrt{36-r^{2}}} 1 \, dz
This evaluates to:
r[z]r36−r2 r \left[ z \right]_{r}^{\sqrt{36-r^{2}}}
=r(36−r2−r) = r \left( \sqrt{36-r^{2}} - r \right)

STEP 4

Substitute the result from Step 1 into the next integral with respect to rr:
The integral becomes:
∫022r(36−r2−r) dr \int_{0}^{2\sqrt{2}} r \left( \sqrt{36-r^{2}} - r \right) \, dr
Distribute rr inside the integral:
=∫022(r36−r2−r2) dr = \int_{0}^{2\sqrt{2}} \left( r\sqrt{36-r^{2}} - r^2 \right) \, dr
This integral can be split into two separate integrals:
=∫022r36−r2 dr−∫022r2 dr = \int_{0}^{2\sqrt{2}} r\sqrt{36-r^{2}} \, dr - \int_{0}^{2\sqrt{2}} r^2 \, dr

STEP 5

Evaluate the first integral:
Let u=36−r2 u = 36 - r^2 , then du=−2r dr du = -2r \, dr or −12du=r dr -\frac{1}{2} du = r \, dr .
Change the limits of integration: when r=0 r = 0 , u=36 u = 36 ; when r=22 r = 2\sqrt{2} , u=28 u = 28 .
The integral becomes:
−12∫3628u du -\frac{1}{2} \int_{36}^{28} \sqrt{u} \, du
Evaluate the integral:
=−12[23u3/2]3628 = -\frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{36}^{28}
=−13((28)3/2−(36)3/2) = -\frac{1}{3} \left( (28)^{3/2} - (36)^{3/2} \right)
=−13(2828−3636) = -\frac{1}{3} \left( 28\sqrt{28} - 36\sqrt{36} \right)

STEP 6

Evaluate the second integral:
∫022r2 dr \int_{0}^{2\sqrt{2}} r^2 \, dr
This evaluates to:
[r33]022 \left[ \frac{r^3}{3} \right]_{0}^{2\sqrt{2}}
=(22)33−0 = \frac{(2\sqrt{2})^3}{3} - 0
=82⋅23 = \frac{8\sqrt{2} \cdot 2}{3}
=1623 = \frac{16\sqrt{2}}{3}

STEP 7

Combine the results from Step 3 and Step 4:
The integral with respect to rr becomes:
−13(2828−3636)−1623 -\frac{1}{3} \left( 28\sqrt{28} - 36\sqrt{36} \right) - \frac{16\sqrt{2}}{3}

STEP 8

Evaluate the final integral with respect to θ\theta:
The integral becomes:
∫02π(−13(2828−3636)−1623) dθ \int_{0}^{2\pi} \left( -\frac{1}{3} \left( 28\sqrt{28} - 36\sqrt{36} \right) - \frac{16\sqrt{2}}{3} \right) \, d\theta
Since the integrand is constant with respect to θ\theta, the integral evaluates to:
(−13(2828−3636)−1623)⋅[θ]02π \left( -\frac{1}{3} \left( 28\sqrt{28} - 36\sqrt{36} \right) - \frac{16\sqrt{2}}{3} \right) \cdot \left[ \theta \right]_{0}^{2\pi}
=(−13(2828−3636)−1623)⋅2π = \left( -\frac{1}{3} \left( 28\sqrt{28} - 36\sqrt{36} \right) - \frac{16\sqrt{2}}{3} \right) \cdot 2\pi
The volume V V is:
V=2π(−13(2828−3636)−1623) V = 2\pi \left( -\frac{1}{3} \left( 28\sqrt{28} - 36\sqrt{36} \right) - \frac{16\sqrt{2}}{3} \right)

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord