Solve ∫ (from 0 - 4 to 2 pi) of int_{sqrt{4-32}}^1dr wrt zdθ

$$\int _ { 0 - 4 } ^ { 2 \pi } \int _ { \sqrt { 4 - 32 } } ^ { 1 } d r d z d \theta$$

Evaluate

$$\text{Indeterminate}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \int _{-4}^{2\pi }dr\mathrm{d}z\mathrm{d}\theta$$

Find the integral of \int _{-4}^{2\pi }dr\mathrm{d}z using the table of common integrals rule \int a\mathrm{d}\theta =a\theta .

$$\int _{-4}^{2\pi }dr\mathrm{d}z\theta$$

Simplify.

$$2dr\left(\pi +2\right)\theta$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$2dr\left(\pi +2\right)\times 1-2dr\left(\pi +2\right)\times 2\left(1-8\right)^{\frac{1}{2}}$$

Simplify.

$$2dr\left(\pi +2\right)$$

Differentiate w.r.t. d

$$\text{Indeterminate}$$