Using cylindrical coordinates, find the volume of the region $D ={y^2+z^2le5+x^2,4x^2+y^2+z^2le25}$
$begingroup$
I want to calculate this volume region, using cylindrical coordinates:
$$D={y^2+z^2le5+x^2,4x^2+y^2+z^2le25}$$
So, I have a hyperboloid and an ellipsoid.
Is it correct to calculate the volume like this:
$$2int_{0}^{2pi}int_{0}^{3}int_{2}^{frac12sqrt{25-r^2}}rdxdrdtheta + 2int_{0}^{2pi}int_{0}^{2}int_{0}^{sqrt{5+x^2}}rdrdxdtheta$$
Is there a more efficient method ?
calculus integration multivariable-calculus definite-integrals
edited Jan 19 at 15:28
Blue
49.5k870158
asked Jan 16 at 17:42
NPLSNPLS
7812
$endgroup$
add a comment |
$begingroup$
I want to calculate this volume region, using cylindrical coordinates:
$$D={y^2+z^2le5+x^2,4x^2+y^2+z^2le25}$$
So, I have a hyperboloid and an ellipsoid.
Is it correct to calculate the volume like this:
$$2int_{0}^{2pi}int_{0}^{3}int_{2}^{frac12sqrt{25-r^2}}rdxdrdtheta + 2int_{0}^{2pi}int_{0}^{2}int_{0}^{sqrt{5+x^2}}rdrdxdtheta$$
Is there a more efficient method ?
calculus integration multivariable-calculus definite-integrals
edited Jan 19 at 15:28
Blue
49.5k870158
asked Jan 16 at 17:42
NPLSNPLS
7812
$endgroup$
add a comment |
$begingroup$
I want to calculate this volume region, using cylindrical coordinates:
$$D={y^2+z^2le5+x^2,4x^2+y^2+z^2le25}$$
So, I have a hyperboloid and an ellipsoid.
Is it correct to calculate the volume like this:
$$2int_{0}^{2pi}int_{0}^{3}int_{2}^{frac12sqrt{25-r^2}}rdxdrdtheta + 2int_{0}^{2pi}int_{0}^{2}int_{0}^{sqrt{5+x^2}}rdrdxdtheta$$
Is there a more efficient method ?
calculus integration multivariable-calculus definite-integrals
edited Jan 19 at 15:28
Blue
49.5k870158
asked Jan 16 at 17:42
NPLSNPLS
7812
$endgroup$
I want to calculate this volume region, using cylindrical coordinates:
$$D={y^2+z^2le5+x^2,4x^2+y^2+z^2le25}$$
So, I have a hyperboloid and an ellipsoid.
Is it correct to calculate the volume like this:
$$2int_{0}^{2pi}int_{0}^{3}int_{2}^{frac12sqrt{25-r^2}}rdxdrdtheta + 2int_{0}^{2pi}int_{0}^{2}int_{0}^{sqrt{5+x^2}}rdrdxdtheta$$
Is there a more efficient method ?
calculus integration multivariable-calculus definite-integrals
calculus integration multivariable-calculus definite-integrals
edited Jan 19 at 15:28
Blue
49.5k870158
asked Jan 16 at 17:42
NPLSNPLS
7812
edited Jan 19 at 15:28
Blue
49.5k870158
asked Jan 16 at 17:42
NPLSNPLS
7812
edited Jan 19 at 15:28
Blue
49.5k870158
edited Jan 19 at 15:28
Blue
49.5k870158
edited Jan 19 at 15:28
Blue
49.5k870158
49.5k870158
asked Jan 16 at 17:42
NPLSNPLS
7812
asked Jan 16 at 17:42
NPLSNPLS
7812
asked Jan 16 at 17:42
NPLSNPLS
7812
7812
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(I don't understand your first integral. The second is o.k.)
Your domain $D$ is a diabolo with the $x$-axis as axis of rotation. It is the intersection of the interior of a one-sheet hyperboloid with an ellipsoid. For the computation we put $y^2+z^2=:rho^2$. The admissible values for $rho$ depend on the variable $x$. When $xgeq0$ we have
$$0leqrholeq rho(x):=sqrt{5+x^2}quad(0leq xleq2),qquad 0leqrholeqrho(x):=sqrt{25-4x^2}quadleft(2leq xleq{5over2}right) .$$
We now cut up $D$ into "infinitesimal discs" of radius $rho(x)$ using planes orthogonal to the $x$-axis. These discs have volume $pirho^2(x)>dx$. It follows that
$${rm vol}(D)=2piint_0^2rho^2(x)>dx+2piint_2^{5/2}rho^2(x)>dx .$$
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
$endgroup$
$begingroup$
I got $30pi$ I think It's right!
$endgroup$
– NPLS
Jan 19 at 15:40
add a comment |
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1 Answer
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$begingroup$
(I don't understand your first integral. The second is o.k.)
Your domain $D$ is a diabolo with the $x$-axis as axis of rotation. It is the intersection of the interior of a one-sheet hyperboloid with an ellipsoid. For the computation we put $y^2+z^2=:rho^2$. The admissible values for $rho$ depend on the variable $x$. When $xgeq0$ we have
$$0leqrholeq rho(x):=sqrt{5+x^2}quad(0leq xleq2),qquad 0leqrholeqrho(x):=sqrt{25-4x^2}quadleft(2leq xleq{5over2}right) .$$
We now cut up $D$ into "infinitesimal discs" of radius $rho(x)$ using planes orthogonal to the $x$-axis. These discs have volume $pirho^2(x)>dx$. It follows that
$${rm vol}(D)=2piint_0^2rho^2(x)>dx+2piint_2^{5/2}rho^2(x)>dx .$$
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
$endgroup$
$begingroup$
I got $30pi$ I think It's right!
$endgroup$
– NPLS
Jan 19 at 15:40
add a comment |
$begingroup$
(I don't understand your first integral. The second is o.k.)
Your domain $D$ is a diabolo with the $x$-axis as axis of rotation. It is the intersection of the interior of a one-sheet hyperboloid with an ellipsoid. For the computation we put $y^2+z^2=:rho^2$. The admissible values for $rho$ depend on the variable $x$. When $xgeq0$ we have
$$0leqrholeq rho(x):=sqrt{5+x^2}quad(0leq xleq2),qquad 0leqrholeqrho(x):=sqrt{25-4x^2}quadleft(2leq xleq{5over2}right) .$$
We now cut up $D$ into "infinitesimal discs" of radius $rho(x)$ using planes orthogonal to the $x$-axis. These discs have volume $pirho^2(x)>dx$. It follows that
$${rm vol}(D)=2piint_0^2rho^2(x)>dx+2piint_2^{5/2}rho^2(x)>dx .$$
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
$endgroup$
$begingroup$
I got $30pi$ I think It's right!
$endgroup$
– NPLS
Jan 19 at 15:40
add a comment |
$begingroup$
(I don't understand your first integral. The second is o.k.)
Your domain $D$ is a diabolo with the $x$-axis as axis of rotation. It is the intersection of the interior of a one-sheet hyperboloid with an ellipsoid. For the computation we put $y^2+z^2=:rho^2$. The admissible values for $rho$ depend on the variable $x$. When $xgeq0$ we have
$$0leqrholeq rho(x):=sqrt{5+x^2}quad(0leq xleq2),qquad 0leqrholeqrho(x):=sqrt{25-4x^2}quadleft(2leq xleq{5over2}right) .$$
We now cut up $D$ into "infinitesimal discs" of radius $rho(x)$ using planes orthogonal to the $x$-axis. These discs have volume $pirho^2(x)>dx$. It follows that
$${rm vol}(D)=2piint_0^2rho^2(x)>dx+2piint_2^{5/2}rho^2(x)>dx .$$
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
$endgroup$
(I don't understand your first integral. The second is o.k.)
Your domain $D$ is a diabolo with the $x$-axis as axis of rotation. It is the intersection of the interior of a one-sheet hyperboloid with an ellipsoid. For the computation we put $y^2+z^2=:rho^2$. The admissible values for $rho$ depend on the variable $x$. When $xgeq0$ we have
$$0leqrholeq rho(x):=sqrt{5+x^2}quad(0leq xleq2),qquad 0leqrholeqrho(x):=sqrt{25-4x^2}quadleft(2leq xleq{5over2}right) .$$
We now cut up $D$ into "infinitesimal discs" of radius $rho(x)$ using planes orthogonal to the $x$-axis. These discs have volume $pirho^2(x)>dx$. It follows that
$${rm vol}(D)=2piint_0^2rho^2(x)>dx+2piint_2^{5/2}rho^2(x)>dx .$$
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
edited Jan 19 at 15:40
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
answered Jan 19 at 15:23
Christian BlatterChristian Blatter
176k8115328
176k8115328
$begingroup$
I got $30pi$ I think It's right!
$endgroup$
– NPLS
Jan 19 at 15:40
add a comment |
$begingroup$
I got $30pi$ I think It's right!
$endgroup$
– NPLS
Jan 19 at 15:40
$begingroup$
I got $30pi$ I think It's right!
$endgroup$
– NPLS
Jan 19 at 15:40
$begingroup$
I got $30pi$ I think It's right!
$endgroup$
– NPLS
Jan 19 at 15:40
add a comment |
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