Calculus · real student question

Evaluate the triple integral of -2y dz dy dx where x runs from -3 to 3, y from 0 to the square root of 9 minus x squared, and z from 0 to 9 minus x squared minus y squared. Give an exact answer.

Question

Evaluate

3309x209x2y2(2y)dzdydx\int_{-3}^{3}\int_{0}^{\sqrt{9-x^2}}\int_{0}^{\,9-x^2-y^2} (-2y)\,dz\,dy\,dx

Enter an exact answer.

Step-by-step solution

  1. Read the limits as a shape before integrating. The xx and yy limits describe x2+y29x^2+y^2\le 9 with y0y\ge 0 — the upper half of the disk of radius 33 — and zz runs from the plane z=0z=0 up to the paraboloid z=9x2y2z=9-x^2-y^2. Recognising the half-disk is what makes the cylindrical cross-check in the last step possible.

  2. Integrate in zz first, because the integrand has no zz. The inner integral is just the integrand times the height of the solid:

    09x2y2(2y)dz=2y(9x2y2)\int_{0}^{9-x^2-y^2}(-2y)\,dz=-2y\,(9-x^2-y^2)

  3. Integrate in yy and watch the two terms merge. Expanding, 2y(9x2)+2y3-2y(9-x^2)+2y^3, so with R2=9x2R^2=9-x^2:

    [(9x2)y2+12y4]09x2=(9x2)2+12(9x2)2=12(9x2)2\Bigl[-(9-x^2)y^2+\tfrac12 y^4\Bigr]_{0}^{\sqrt{9-x^2}}=-(9-x^2)^2+\tfrac12(9-x^2)^2=-\tfrac12(9-x^2)^2

    The upper limit squares to exactly 9x29-x^2, which is why the two powers of RR combine so cleanly.

  4. Integrate in xx using the evenness of the integrand. Since (9x2)2=8118x2+x4(9-x^2)^2=81-18x^2+x^4 is even,

    1233 ⁣ ⁣(8118x2+x4)dx=03 ⁣ ⁣(8118x2+x4)dx-\frac12\int_{-3}^{3}\!\!\bigl(81-18x^2+x^4\bigr)dx=-\int_{0}^{3}\!\!\bigl(81-18x^2+x^4\bigr)dx
    =[81x6x3+x55]03=(243162+2435)=6485=-\Bigl[81x-6x^3+\tfrac{x^5}{5}\Bigr]_0^3=-\left(243-162+\frac{243}{5}\right)=-\frac{648}{5}

  5. Confirm the value in cylindrical coordinates. With y=rsinθy=r\sin\theta, dV=rdzdrdθdV=r\,dz\,dr\,d\theta, the same solid is 0r30\le r\le 3, 0θπ0\le\theta\le\pi, 0z9r20\le z\le 9-r^2:

    0π ⁣ ⁣03(2rsinθ)(9r2)rdrdθ=2(0π ⁣sinθdθ)(03 ⁣r2(9r2)dr)\int_0^{\pi}\!\!\int_0^{3}(-2r\sin\theta)(9-r^2)\,r\,dr\,d\theta=-2\left(\int_0^{\pi}\!\sin\theta\,d\theta\right)\left(\int_0^{3}\!r^2(9-r^2)\,dr\right)

    The two factors are 22 and 812435=162581-\tfrac{243}{5}=\tfrac{162}{5}, giving 221625=6485-2\cdot 2\cdot\tfrac{162}{5}=-\tfrac{648}{5} — the same exact value from a completely different route.

  6. Interpret the sign. The whole solid sits in y0y\ge 0 and the integrand 2y-2y is 0\le 0 there, so a negative answer is expected; a positive result would immediately signal a sign slip.

Answer

6485-\frac{648}{5}

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